Physics Book - User contributions [en]

Charged Capacitor

2019-08-21T20:48:44Z

Parkercoye: /* Difficult */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 

===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png|300px|thumb|Figure 7: Difficult problem]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-21T20:48:23Z

Parkercoye: /* Middling */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png|200px|thumb|Figure 6: Middling problem #2]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 

===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-21T20:47:30Z

Parkercoye: /* Connectedness */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==
ADD MORE INFO ABOUT CONNECTEDNESS HERE

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-21T20:47:20Z

Parkercoye: /* History */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==
ADD MORE INFO ABOUT HISTORY HERE

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Spherical Shell

2019-08-21T20:46:34Z

Parkercoye: /* Observation Location Within the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because

[[File:ShellHollow2.JPG|400px|thumb|Figure 2: gaussian surface inside of a spherical shell]]

the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:46:15Z

Parkercoye: /* Observation Location Inside the Hollow Portion of the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because

[[File:ShellHollow2.JPG|400px|thumb|Figure 2: gaussian surface inside of a spherical shell]]

the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:45:36Z

Parkercoye: /* Observation Location Outside the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:45:10Z

Parkercoye: /* Observation Location Within the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg|400px|thumb|Figure 3: gaussian surface inside spherical shell]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:44:32Z

Parkercoye: /* Observation Location Inside the Hollow Portion of the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:44:22Z

Parkercoye: /* Observation Location Inside the Hollow Portion of the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:43:51Z

Parkercoye: /* Observation Location Outside the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG|400px|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:43:14Z

Parkercoye: /* Observation Location Outside the Spherical Shell: */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG|thumb|Figure 1: a gaussian surface outside of a spherical shell]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:41:30Z

Parkercoye: /* A Computational Model */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

==A Computational Model==
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Charged Spherical Shell

2019-08-21T20:41:22Z

Parkercoye: /* A Mathematical Model */

==The Main Idea==

A charged spherical shell is referring to the idea that there is a solid object that can be defined as the space between two concentric spheres that has a uniformly distributed charge, in other words, a hollow sphere that has some thickness. Charged objects create electric fields and this electric field depends on the object's shape, charge, and the distance to the observation location. In the case of a charged spherical shell, if the observation location is within the hollow portion of the shell (distance less than the inner radius of the spherical shell) the electric field is zero. If the observation location is outside of the shell (distance greater than the outer radius of the sphere) then the equation to calculate the electric field of a point charge can be applied. If the observation location is within the spherical shell itself (distance r between inner radius a and outer radius b) then the electric field of the shell can be found using the charge contained within the radius r instead of the total charge.

==A Mathematical Model==

When considering how to calculate the electric field of a charged spherical shell you first need to identify where the observation location is, and then follow the guidelines below.

Observation location outside of the spherical shell: <math>\vec{E}_{sphere}=\frac{1}{4\pi E_0}\frac{Q}{r^2}\widehat{r}</math>

Observation location within the spherical shell: <math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}\widehat{r}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q\widehat{r}</math>

Observation location inside of the hollow portion of the spherical shell: <math>\vec{E}_{sphere}=0</math>

===A Computational Model===
Linked below is a VPython simulation of the electric field of a charged spherical shell. The sphere is modeled in this simulation as a series of rings stacked on top of one another with varying radii. The white arrows represent the electric field at any given point, and the yellow trail represents the theoretical path of an electron orbiting the sphere.
====Observation Location Outside the Spherical Shell:====

If the observation location is outside of the shell, the electric field produced mirrors that of a point charge, due to the shape and charge distribution of the charged spherical shell. Say the shell is located at the origin, and the observation location is on the x-axis. The direction of the electric field produced by the shell at the observation location is in the x direction. This is because all of the other electric field vectors with y and x components cancel out in the y direction, leaving only the electric field in the x direction. The same logic would be used if the observation location was on any of the axes. For example, if the observation location had a unit vector of <1,1,0>, then the electric field would have components in the x and y directions, and their magnitudes would be whatever the value of the electric field was found to be multiplied by 1, since both the x and y components of the unit vector have values of 1.

See this [https://trinket.io/embed/glowscript/c7b5085ea4?start=result trinket] for an example of an electric field of a spherical shell with observation locations outside of the shell.

The idea that the electric field produced by a spherical shell can be modeled with a point charge can be explained through [[Gauss's Law]]. Gauss's Law allows one to find the electric field for symmetric charge distributions. Gauss's Law states <math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>.

Gauss's Law considers electric flux through the surface, in which the Electric Field is perpendicular to surface. Therefore, any Electric Field that is tangential to the surface is not considered when determining the charge on a spherical shell.

[[File:ShellOutside.JPG]]

If a sphere of a radius equal to the distance between the observation location and the center of the shell is used for the gaussian surface,then the area can be replaced as follows: <math>E = \frac{Q_{enclosed}}{\varepsilon_0 4\pi r^2}</math>.

In other words, <math>E = \frac{1}{ 4\pi \varepsilon_0} \frac{Q_{enclosed}}{r^2}</math> which is the equation for an electric field of a charged particle.

====Observation Location Inside the Hollow Portion of the Spherical Shell:====

If the observation location is anywhere inside of the spherical shell, then the electric field is zero. If there are positive charges distributed equally around a spherical shell, then those charges contribute to the net electric field. Suppose we choose a point inside the sphere, point A. The contributions of the different charges in the spherical shell cancel each other out. The closer a charge is to point A inside the sphere, the larger its contribution to the electric field at point A. This is because the distance from an observation location to a charge (r) is inversely proportional to the magnitude of the electric field the charge produces. If point A is closer to one side of the sphere, then that means there are more charges on the other side of the sphere that affect the observation location. The extra magnitude from the greater number of charges on the far side cancel out the increased magnitude of the close side. This causes the electric field to be zero at point A inside the sphere, and this can be generalized to any point inside the sphere.

This concept can also be shown through [[Gauss's Law]].

<math>\frac{Q_{enclosed}}{\varepsilon_0} = \oint_C E\bullet dA</math>

Using a gaussian surface of a sphere of radius r where r<a0 = \oint_C E\bullet dA</math> and thus the electric field through the surface is also 0.

'''IMPORTANT:'''
It is very important to remember that the electric field inside a spherical shell is NOT always zero. Other charges that exist may contribute to the electric field inside the spherical shell, only the electric field that is being caused by the uniformly distributed charges within the spherical shell is zero.

====Observation Location Within the Spherical Shell:====

The electric field at an observation location within the spherical shell itself is dependent on the portion of the shell that is closer to the center of the shell than the observation location. In other words, the electric field is dependent on the distance from the observation location to the center of the spherical shell, the outer radius of the shell, and the inner radius of the shell.
If there is a uniformly charged spherical shell of total charge Q with an outer radius of b, an inner radius of a, the electric field at an observation location radius r away from the center of the shell (a<r<b) can be found as follows:

The charge contained within radius r: <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

Using [[Gauss's Law]] the electric field through the gaussian surface (a sphere of radius r):

[[File:ShellWithin.jpg]]

<math>E = \frac{Q_1}{\epsilon_0*4\pi r^2}</math>

There is no electric field contributed by the portion of the shell that is of a radius greater than r, as the observation location would then be inside the hollow section of this larger shell and the electric field would be 0. Thus the electric field at the observation location is only due to the portion of the shell that is less than a distance r away from the center of the shell.

==Examples==

===Simple===
A spherical shell of charge with a radius of 5 is located at the origin and is uniformly charged with q=+2. What is the electric field produced from the spherical shell at x=2?

<math>E = 0</math>

===Middling===
A spherical shell with a radius of 1m is located at the origin and is uniformly charged with q=+6e-8. What is the electric field produced from the spherical shell at (6,3,2)?

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\widehat{r}</math>

<math>E_{sphere} = \frac{1}{4\pi \epsilon_0}\frac{6*10^{-8}}{7^2}<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

<math>E_{sphere} ≈\ (11.02)<\frac{6}{7},\frac{3}{7},\frac{2}{7}></math>

===Difficult===
A spherical shell of charge Q = +5e-10 with an outer radius of 1m and an inner radius of .8m is located at the origin and uniformly charged. What is the electric field produced by the spherical shell at (.9,0,0)?

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{Q_1}{r^2}</math> where <math>Q_1 = \frac{r^3-a^3}{b^3-a^3}Q</math>

<math>Q_1 = \frac{.9^3-.8^3}{1^3-.8^3}((5)10^{-10})</math>

<math>\vec{E}_{sphere} = \frac{1}{4\pi\epsilon_0}\frac{\frac{.9^3-.8^3}{1^3-.8^3}(5*10^{-10})}{.9^2}</math>

<math>E_{sphere} ≈\ <2.74489,0,0></math>

=== Step Problem===
A conducting spherical shell has an inner radius of a, outer radius of b (a<r<b). It also has a positive point charge of +Q at the center. The total charge on the shell is -4Q.

Find: (1) Electric Field r >b from center. (2) Surface Charge Density on the Inner Surface (3) Surface Charge Density on the Outer Surface

Solution:
(1)Since this is a conducting shell the Electric Field is pointing inward then the sign is negative. Also the net charge is -4Q+Q=-3Q, this value replaces Q enclosed in the aforementioned section, Computational Model - Outside of the Shell. Therefore the the Electric Field can be noted as: <math>E = \frac{-1}{ 4\pi \varepsilon_0} \frac{3*Q_{enclosed}}{r^2}</math>

(2)Surface density is the ratio of surface charge to the surface area of the shell, which is a circle. Therefore the surface charge density for the inner surface of a shell can be noted as (using only charge on inside, -Q): <math>E = \frac{-Q_{(inner)}}{4\pi{r^2}}</math>

(2)The surface charge density for the outer surface of a shell can be noted as (using net, -3Q): <math>E = \frac{-3Q_{(inner)}}{4\pi{r^2}}</math>.

==Connectedness==

This topic is related to electric fields and the effects that electric fields can have on other objects. For example, electric fields can have effects on humans! The body's voltage can be increased, currents can be induced by the body, and electric charges can buildup on the surface of peoples' skin which is why they feel a tingling sensation when exposed to electric fields (such as from standing under a high voltage power line). This tingling is felt starting from voltages of 1,000 volts per meter. At that same voltage, there are microdischarges when a person touches something made of metal.

This topic is also very closely tied to [[Gauss's Law]] as can be seen in the computational model. The electric field can always be found by using Gauss's Law and the charge contained within a certain radius.

==History==

The electric field from a point charge was discovered by Charles Augustin de Coulomb, a French physicists. Coulomb's law was published in 1784. The law states that the electric field from a point charge is inversely proportional to the distance between the charged particle and the observation location. It also states that if the charge creating the electric field is positive, then the electric field will point radially outward. However, if the particle creating the field is negatively charged, then the electric field will point radially inward.

== See also ==

[[Electric Force]] One application of electric fields due to point charges deals with finding electric force

[[Electric Field]] More general ideas about electric fields 

[[Gauss's Law]] More information behind why Gauss's law works and why it can be applied to spherical shells

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

http://www.hydroquebec.com/fields/corps-humain.html

http://physicstasks.eu/1531/field-of-charged-spherical-shell

https://www.youtube.com/watch?v=wOPdRcZY-bY

==References==

Note: all images are original and were produced by the author

[[Category:Which Category did you place this in?]]

Field of a Charged Ball

2019-08-21T20:40:52Z

Parkercoye: /* The Main Idea */

==The Main Idea==
In this section, we will discuss the electric field of a solid sphere. Such a sphere has charge distributed throughout the volume (rather than only on the surface), and can be modeled by several layers of concentric, charged spherical shells. Calculating the electric field both outside and inside the sphere will be addressed.

[[File:Wiki_pic.JPG|400px|right|thumb|Figure 1: A sphere with uniformly distributed charge: note that this can be thought of as infinitely thin concentric charged shells]]
==A Mathematical Model==

First we must determine the relationship between r, the radius of the observation point from the center of the sphere, and R, the radius of the sphere itself.

Here, it is necessary to determine whether the observation point is outside or inside the sphere.

If r>R, then we are outside the sphere. As a result, the sphere can be simply treated as if the sphere were a point charge located at the center of the sphere. Hence, the electric field at any point outside of the radius of the sphere can be calculated using the formula for the electric field of a point charge.

<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>
when r>R, and R is the radius of the sphere.

However, when r<R, the observation location is inside of the sphere and the sphere overall must be thought of as infinitely many concentric charged shells inside of each other. All of the shells with a radius larger than that of the observation location do not contribute to the electric field at the observation location. This phenomena is because the electric field produced by these larger shells cancels out at any given point inside of itself. As a result only the shells smaller than the radius of the observation location need to be accounted for.
To find <math>\vec E_{net} </math>, add the contributions to the electric field from the inner shells.
After adding the contributions of each inner shell, you should have an electric field equal to:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{\Delta Q}{r^2}</math>

We find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

We find that:

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

The charge inside the sphere is proportional to r. When r=R, we again treat the sphere as a point charge located at the center of the sphere.

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^2} </math>

==A Computational Model==

This demonstration using VPython shows the field of a charged ball with a fixed radius. Try changing the radius to see what happens!

https://trinket.io/glowscript/48f6efd07a

# GlowScript 1.1 VPython
## constants
oofpez = 9e9
qproton = 1.6e-19
scalefactor = 2e-20
Radius =3.5
r=3

## objects
particle = sphere(pos=vector(0,0,0), opacity =0.5, radius=Radius, color=color.blue)
## initial values
obslocation = vector(r,0,0)
##calculate position vector
rvector = obslocation - particle.pos
arrow1 = arrow(pos=particle.pos, axis=rvector, color=color.green)
rmag=mag(rvector)

##colored arrow

if r<= Radius:
E1= oofpez*(qproton)/(Radius)**3 *rmag* vector(1,0,0)
E2= oofpez*(qproton)/(Radius)**3 *rmag* vector(-1,0,0)
E3= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,1,0)
E4= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,-1,0)
E5= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,1)
E6= oofpez*(qproton)/(Radius)**3 *rmag* vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)
if r>Radius:
E1= oofpez*(qproton)/(rmag)**2 *vector(1,0,0)
E2= oofpez*(qproton)/(rmag)**2 *vector(-1,0,0)
E3= oofpez*(qproton)/(rmag)**2 * vector(0,1,0)
E4= oofpez*(qproton)/(rmag)**2 *vector(0,-1,0)
E5= oofpez*(qproton)/(rmag)**2 *vector(0,0,1)
E6= oofpez*(qproton)/(rmag)**2 * vector(0,0,-1)
ea1 = arrow(pos=vector(rmag,0,0), axis=E1*scalefactor, color=color.orange)
ea2 = arrow(pos=vector(-rmag,0,0), axis=E2*scalefactor, color=color.orange)
ea3 = arrow(pos=vector(0,rmag,0), axis=E3*scalefactor, color=color.orange)
ea4 = arrow(pos=vector(0,-rmag,0), axis=E4*scalefactor, color=color.orange)
ea5 = arrow(pos=vector(0,0,rmag), axis=E5*scalefactor, color=color.orange)
ea6 = arrow(pos=vector(0,0,-rmag), axis=E6*scalefactor, color=color.orange)

==Examples==

===Simple===
[[File:Approx_point.png|200px|right|thumb|Figure 2: This is an example of applying the equation for finding the electric field of a point charge]]
1. Describe the pattern of the electric field of charged ball from far away.

ANS: If we observe a charged ball from far away, we can say that r>R. Because we are outside of the sphere, the ball can essentially be viewed as a point charge. Thus the electric field can be defined by:
<math>\vec E=\frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} \hat r</math>

===Middling===
[[File:Concentric_sphere.png|200px|right|thumb|Figure 3: This is very similar to problem 1 and involves calculating the field of a point charge.]]
A sphere is charged throughout it's volume with a charge of Q= 6e-5 C. The radius of the this sphere is R=10. Find the electric field created by a sphere of radius r=4.

Step 1: Cut up the sphere into shells.

step 2: We know that r<R.

Next find <math> \Delta Q </math>:

<math> \Delta Q = Q \frac{\text {volume of inner shells}}{\text {volume of sphere}} = Q \frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3}</math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2}\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r = 9e9 * \frac{6*10^{-5} C}{(10m)^3}*4m = 2160 N/C </math>

===Difficult===
A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge −e. (The actual charge density in the ground state is nonuniform.)

[[File:circle_outline.png|200px|thumb|left|Figure 4: This circle represents a hydrogen atom. Use the radius and total charge to find alpha ]]

For the uniform-density model, calculate the polarizability α of atomic hydrogen in terms of R. Consider the case where the magnitude E of the applied electric field is much smaller than the electric field required to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud

(s « R in the diagram). (Use the following as necessary: R and ε0.)

Useful equations:

<math> \vec p = \alpha *E</math>

<math> \vec p = Q*s </math>

Assume an applied electric field of strength E. This electric field polarized the hydrogen atom. Now there is a spherical charge of radius s. Use the volume ratio, and then use the useful equations to find <math> \alpha </math>

<math>\vec E = \frac{1}{4 \pi \epsilon_0}\frac{e}{s^2}\frac{\frac{4}{3} \pi s^3}{\frac{4}{3} \pi R^3} = \frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s </math>

<math>\alpha = \frac{e*s}{\frac{1}{4 \pi \epsilon_0}\frac{e}{R^3}s} = 4\pi \epsilon_0 R^3 </math>

==Looking Ahead, Gauss's Law==

Later in this wikibook, you will learn about Gauss's Law. This will make calculating the electric field easier.

The formula for Gauss's Law is:

<math>\oint \vec E \bullet \hat n dA = \frac{1}{\epsilon_0}\Sigma Q_{\text {inside the surface}} </math>

Using this, one can calculate the the electric field when r<R for a solid sphere charged throughout.

<math>\vec E * 4\pi*r^2= \frac{1}{\epsilon_0}Q\frac{\frac{4}{3} \pi r^3}{\frac{4}{3} \pi R^3} </math>

<math> \vec E = \frac{1}{4 \pi \epsilon_0}\frac{Q}{R^3}r </math>

==History==
ADD MORE ABOUT HISTORY HERE

==Connectedness==
It may be particularly useful to discuss real-life applications of a charged solid sphere. Two examples stem from the structure of an atom. The nucleus of an atom is packed very tightly so that we can consider the charge to be uniformly distributed. The electron cloud also can be viewed as a packed spherical region of charged. Of course the radii of these structures are very small; radii are about 10e-15 m and 10e-10 m for nuclei and electron clouds respectively!

== See also ==

===Further reading===

http://www.physicsbook.gatech.edu/Point_Charge

http://www.physicsbook.gatech.edu/Electric_Dipole

http://www.physicsbook.gatech.edu/Gauss%27s_Flux_Theorem

==References==

This section contains the the references you used while writing this page

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

https://www.physicsforums.com/threads/calculate-the-polarizability-a-lpha-of-atomic-hydrogen-in-terms-of-r.339994/

Matter and Interactions, 4th Edition: 1-2

Charged Capacitor

2019-08-20T20:12:19Z

Parkercoye: /* Examples */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==
===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png|200px|thumb| Figure 5: Middling problem ]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:11:22Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 

 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:11:01Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|200px|thumb|Figure 4: Problem 1.2 diagram]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:10:29Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|200px|right|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|300px]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:10:19Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|200px|center|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|300px]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:10:07Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|300px|left|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|300px]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:09:49Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|300px|right|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|300px]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:09:30Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|300px|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png|300px]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:09:11Z

Parkercoye: /* Simple */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|300px|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 

===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:09:02Z

Parkercoye: /* Examples */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png|400px|thumb|Figure 3: Problem 1 diagram]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:07:42Z

Parkercoye: /* Derivation */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|250px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:07:33Z

Parkercoye: /* Gauss's Law */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|400px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|250px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:07:06Z

Parkercoye: /* Gauss's Law */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|400px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|300px|center|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:06:52Z

Parkercoye: /* Gauss's Law */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|400px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|300px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:06:42Z

Parkercoye: /* Gauss's Law */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|400px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png|400px|thumb| Figure 2: Diagram of Gauss' Law within a charged capacitor]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T20:05:25Z

Parkercoye: /* Derivation */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png|400px|thumb|Figure 1: Diagram of a capacitor]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 

====Gauss's Law====
[[File:CapacitorII.png]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T19:59:46Z

Parkercoye: /* The Main Idea */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s which stores electric charge. The effect of a capacitor is capacitance, which represents how an electric charge changes with respect to the electric potential.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 
====Gauss's Law====
[[File:CapacitorII.png]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T17:02:51Z

Parkercoye: /* Further reading */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 
====Gauss's Law====
[[File:CapacitorII.png]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

==History==

==Connectedness==

==Further reading==

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T17:02:15Z

Parkercoye: /* A Mathematical Model */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 
====Gauss's Law====
[[File:CapacitorII.png]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

==A Computational Model==

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Capacitor

2019-08-20T17:02:03Z

Parkercoye: /* A Mathematical Model */

==The Main Idea==

A capacitor is when two uniformly, but oppositely (-Q and +Q), charged metal plates are held very close to each other with a separation of s.
 This page is dedicated to understanding and calculating the electric field of a capacitor through definition, mathematical models, computational models, and example problems.

==A Mathematical Model==

The mathematical model to the electric field of a charged capacitor (near the center of the capacitor) is <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>, where '''Q''' is the magnitude of the plate charges and '''A''' is the area of each plates. The direction is perpendicular to the plates.
 The fringe field (field located near the center of the disks but right outside of the plates) is <math>E_{fringe} \approx {\frac{Q/A}{2{\epsilon}_0}} (\frac{s}{R})</math>
 
====Derivation====
[[File:Capacitor111.png]]
 Take the origin at the surface of the left plane, with the z-axis running to the right. We assume that each disk has a uniformly charge density (s <math>\ll</math> R).
 Then, the contribution of the negative capacitor is <math>E_- \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}]</math> (to the left) and the positive capacitor is <math>E_+ \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}]</math> (to the left).
 If we add up the contributions, <math>E_{total} \approx {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{z}{R}] + {\frac{Q/A}{2{\epsilon}_0}} [1-\frac{s-z}{R}] \approx {\frac{Q/A}{{\epsilon}_0}} [1-\frac{s/2}{R}]</math>. Since s <math>\ll</math> R, <math>E \approx {\frac{Q/A}{{\epsilon}_0}}</math>.
 
====Gauss's Law====
[[File:CapacitorII.png]]
 Say <math>\sigma</math> is the surface charge density, and the area on each side of the cylinder is A.
 Since Gauss's Law is <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math>, <math>\Phi_{left}</math> = 0, <math>\Phi_{circular sides}</math> = 0, and <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math>; while <math>\vec{E}_{left} = 0</math>, <math>\vec{E}_{circular sides} \perp \vec{A}</math>.
 Thus, EA = <math>\Phi = \lmoustache \vec{E} \cdot d\vec{A}</math> = <math>\Phi_{right} = \frac{Q_{enclosed}}{\varepsilon_0}</math> = <math>\frac{\sigma A}{\varepsilon_0}</math>. Therefore, E = <math>\frac{\sigma}{\varepsilon_0}</math>

===A Computational Model===

Uniformly charged capacitors can be further explored through [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab PhET Interactive Simulations]. Here, the user can explore how a capacitor works by changing the size of capacitors and add different objects such as dielectrics to observe how they affect capacitance.
 
Charged capacitors can also be visualized through this vpython code created on the website [https://trinket.io/glowscript/3088e75439 Teach hands-on with GlowScript]

==Examples==

Some examples of capacitor problems:

===Simple===

'''<math>\bullet</math> If the plate separation, s, for a capacitor is <math>3 \times 10^{-3}</math>m, determine the area of the plates if the capactitance is 3 F.'''
 
[[File:asdfg.png]]
 Since <math>C = \frac{\varepsilon_{0}A}{d}</math>,
 <math>A = \frac{Cd}{\varepsilon_{0}}</math>
 <math>A = \frac{3 F\times (3\times10^{-3}) m}{8.85\times10^{-12}C^2 N^{-1} m^{-2}}</math> = <math>1.02\times10^{9} m^2</math>.
 
'''<math>\bullet</math> Determine the amount of charged on one side of a capacitor with the capacitance of <math>2\times10^{-6} F</math> when the capacitor is connected to a 12 V battery.'''
[[File:asdf.png]]
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V</math>
 <math>Q = (2\times10^{-6} F\times12 V) = 2.4\times10^{-5} C</math>.
 
===Middling===

'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
 
[[File:qwer.png]]
 The capacitors are connected in series in the top and bottom row, so in each row, the total capacitance is <math>\frac{1}{4\mu} F + \frac{1}{4\mu} F = 2\mu</math> F. The two rows are connected in parallel, so thus, the effective capacitance is <math>2\mu F + 2\mu F = 4\mu</math> F.
 To find the charge on each of the 2<math>\mu</math> F capacitor, the equation <math>C = \frac{Q}{V}</math> is used.
 Since <math>C = \frac{Q}{V}</math>,
 <math>Q = C \times V = 2\mu F\times 20 V = 40\mu C</math>
 To find the voltage on each rows containing the two 4<math>\mu</math> F capacitor,
 Since <math>C = \frac{Q}{V}</math>,
 <math>V = \frac{Q}{C} = \frac{40 \mu\frac{C}{V}}{2\mu F}</math> = 10 V.
 
'''<math>\bullet</math> Evaluate the circuit and determine the effective capacitance and charge and voltage across each capacitor.'''
[[File:qwertt.png]]
 SInce all three capacitors are connected in series in the bottom row, the bottom row's total capacitance will be <math>\frac{1}{6\mu} F + \frac{1}{6\mu} F + \frac{1}{6\mu} F = 2\mu</math> F. The top row and the bottom row are connected in parallel, so the effective capacitance is <math>4\mu F + 2\mu F = 6\mu F</math>.
 Since <math>C = \frac{Q}{V}</math>,
 the charge across the <math>4\mu</math> F capacitor is <math>Q = C \times V = 4\mu F\times 100 V = 400 \mu C</math>,
 and the charge across the row containing the three <math>6\mu</math> capacitors is <math>Q = C \times V = 2\mu F\times 100 V = 200 \mu C</math>.
 The voltage across the <math>6\mu</math> capacitor is <math>V = \frac{Q}{C} = \frac{200\mu C}{6\mu F} = 33.3 V</math>
 
===Difficult===

'''<math>\bullet</math> A proton traveling in the -x direction approaches a capacitor consisting of two large, oppositely charged plates. The circular plates are 0.04 m apart, and each has a radius of 1.7 m. The proton enters the capacitor through a small hole in location A at the right side of plates at <math>2.5 \times 10^{4} \frac{m}{s}</math> and exists through the hole at location B with a speed of <math>7.9 \times 10^{4} \frac{m}{s}</math>.'''
 
[[File:zxcv.png]]
 
 First, we have to find the potential difference between B and A <math>(V_B - V_A)</math>.
 Since this is a closed system, <math>\Delta K + \Delta U = 0</math>. Then,
 <math>\Delta U = -\Delta K = -\frac{1}{2} m ({v_f}^2 - {v_i}^2) = -\frac{1}{2} (1.7e-27)[{7.9e4}^2 - {2.5e4}^2] = -4.77e-18 J</math>
 <math>\Delta U = q\Delta V \Leftrightarrow \Delta V = \frac{1}{q} \Delta U = \frac{-4.77e-18 J}{1.6e-19 C} = -29.8 \frac{J}{C}</math>.
 Since <math>E_{capacitor} = \frac{Q/A}{\varepsilon_0} = \frac{Q}{\pi R^2 \varepsilon_0}</math>, and for a uniform electric field, \Delta V = Ed<math></math>,
 <math>\Delta V = \frac{Q}{\pi R^2 \varepsilon_0} \Leftrightarrow Q = \frac{\Delta V \pi R^2 \varepsilon_0}{d} = \frac{29.8 V \pi ({1.7}^2 m^2) (8.85e-12 C^2 N^{-1} m^{-2})}{0.04 m} = 5.99e-8 C</math>.

===Further reading===

Capacitors can also be read in [http://www.freebookcentre.net/Physics/Electricity-Magnetism-Books.html Free Book Centre], where there are free electricity and magnetism books.

===External links===

Charged capacitors are further explained in websites such as:
 <math>\bullet</math> [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html Hyperphysics]
 <math>\bullet</math> [https://www.khanacademy.org/science/physics/circuits-topic/circuits-with-capacitors/v/capacitance Khan Academy]
 <math>\bullet</math> [http://physics.info/capacitors/ Physics.info]

==References==

Chabay, R. W. (2015). Matter and interactions: Electric and magnetic interactions. Wiley.
 
D. (2013). Gauss' Law Explained and Parallel Plate Capacitor Worked Examples | Doc Physics. Retrieved April 17, 2016, from https://www.youtube.com/watch?v=hnt07JpNE1E
 
Get Ready. Be Prepared. Understand the Big Ideas. (n.d.). Retrieved April 17, 2016, from http://www.physics-prep.com/index.php/practice-problems-capacitors/43-physics-2-unit-1/1437-practice-problems-capacitance-solutions

[[Category: Electric Fields]]

Charged Disk

2019-08-20T17:01:39Z

Parkercoye: /* External Links and References */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|center|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Note: all images were created by author

Charged Disk

2019-08-20T17:01:26Z

Parkercoye: /* History */

Charged Disk

2019-08-20T17:01:13Z

Parkercoye: /* History */

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png|400px|thumb|Figure 5: a capacitor]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T17:00:27Z

Parkercoye:

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T17:00:05Z

Parkercoye: /* Middling */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 3: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:59:56Z

Parkercoye: /* Easy */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 2: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 4: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:59:39Z

Parkercoye: /* Difficult */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 4: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Difficult problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:59:28Z

Parkercoye: /* Difficult */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 4: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png|300px|thumb|Figure 4: Hard problem #3 diagram]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:53:56Z

Parkercoye: /* Middling */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|300px|thumb|Figure 4: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:53:47Z

Parkercoye: /* Middling */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png|400px|thumb|Figure 4: Middling problem #2]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:53:21Z

Parkercoye: /* Easy */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|300px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:53:14Z

Parkercoye: /* Easy */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|400px|thumb|Figure 3: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:53:06Z

Parkercoye: /* A Computational Model */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|400px|thumb|Figure 4: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:52:56Z

Parkercoye: /* Easy */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==
[[File:circle_5.png|300px|thumb|Figure 2: model of charged disk with concentric circles]]

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png|400px|thumb|Figure 4: Easy problem #1]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

[[Category:What category did you place this in?]]

Charged Disk

2019-08-20T16:52:33Z

Parkercoye: /* A Computational Model */

The electric field of a uniformly charged disk can be evaluated in several ways, and in several scenarios. We will cover how to find the electric field of a uniformly charged disk, and how this can also apply to capacitors. (Shubham Shah)

==The Main Idea==

The electric field of a charged disk can be imagined as the total electric field produced from each point on the disk. We can relate this the the Law of Superposition, but considering the disk as a collection of rings, which will lead to a better understanding of the electric feild produced by the disk. This is especially important because two oppositely charged metal disks collectively are known as a [[capacitor]], a concept seen in several places in physics and the real world.

==A Mathematical Model==

Before we begin our calculations, take a look at this image of a uniformly charged disk:

[[File:circle_6.png|400px|thumb|Figure 1: diagram of a uniformly charged disk]]

This image shows a visual representation of a disk. It should look pretty familiar. In fact, the shape of a disk is simply an extension of a [http://physicsbook.gatech.edu/Charged_Ring ring]. Think of it as infinitely many uniformly charged rings. Thinking of a disk this way will help to understand the equation of the electric field for a uniformly charged disk.

Recall the equation for the electric field of a uniformly charged ring:

<math>\ E= \frac{1}{4π\epsilon_0}\frac{qz}{(R^2+z^2)^{3/2}}</math>

This equation will tell you the electric field of a uniformly charged ring at any observation location z. To apply this equation to a disk, integration will be involved. The integration variable in this case should be the radius of the ring <math> r </math>, as that will change with the infinitely many concentric rings you have. However, <math> r </math> is nowhere to be found in the equation. But if the radius of each ring is different, what gets affected as a result?

<math>\
dq = Q\frac{area of ring}{area of disk} = \frac{π((r+dr)^2-r^2)}{πR^2} = \frac{π(r^2 - r^2 +2r*dr+ dr^2)}{πR^2} </math>

We assume that <math> dr^2 </math> is essentially zero compared to the non-squared differential term.

<math> dq= Q\frac{2πrdr}{πr^2}</math>

Here you see that the charge of each ring is different due to the changing radius. The charge changes by a factor of the area of the ring, <math> 2πrdr </math>, (if you roll out the ring, it is a rectangle with height <math> dr </math> and width <math> 2πr </math>) divided by the area of the disk, <math> πr^2 </math>. Plugging that in for the q variable in our integration, and then cancelling out some variables, will allow us to solve for the electric field of a disk:

<math>\ dE = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}\frac{zrdr}{(R^2+z^2)^{3/2}}</math>

Now we take the integral of this equation, and the result is this:

<math>\ E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

Thus, we have the equation that finds us the exact electric potential of a uniformly charged disk. You can still make approximations to make the formula simpler, however. If your observation location z is very close to the disk but not touching it, and is also significantly smaller than the radius of the disk (i.e. <math> 0 << z << R </math>), the equation becomes this:

<math>\ E = \frac{Q}{2\epsilon_0A} </math>

This is the most commonly seen form of this equation, and becomes very important when applied to the case of capacitors. A key aspect of the field in these situations is the fact that it's constant in space between two charged plates. This unique aspect of capacitors, which does not occur for any other type of charged geometry, allows for several novel applications.

==A Computational Model==
[[File:circle_5.png|300px|thumb|Figure 2: model of charged disk with concentric circles]]

The writers of the textbook Matter and Interactions 3rd Edition, have some great examples of VPython [http://www.phy.uct.ac.za/demonline/virtual/scripts/15_E_disk.py code] that show, in computational form, the electric field of a uniformly charged disk. To see the code in action, take a look at [https://trinket.io/glowscript/3088e75439 this] or copy the code into your VIDLE shell. Enjoy!

Also took a look at [https://phet.colorado.edu/en/simulation/legacy/capacitor-lab this] java applet for a lab simulation of a capacitor

==Examples==

===Easy===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:circle_4.png]]

This question is important, as this situation can come up a lot when solving physics problem. Remember that in the special case <math> 0 << z << R </math>, you are allowed to use the approximate formula. But if that second condition, <math> z << R </math>, is not true, then you must use the exact formula (listed above). Or, if <math> 0 << z </math>, you also must use the exact formula, as then the observation location is relatively far from the disk. So it is important to look at your <math> z </math> value when you decide which equation to use when finding the electric field of a uniformly charged disk.

===Middling===

[[File:circle_3.png]]

A thin plastic disk, with it's center on the origin, and a of radius 0.4 m is uniformly charged with charge <math> Q = -4 * 10^{-7} C</math>. What is the electric field of this disk at a location <math> <0,0,0.04> </math>m?

This is a a pretty simple problem, as <math> 0 << z << R </math>, so you can use the approximate formula:

<math> \ E = \frac{4 * 10^{-7}}{2\epsilon_0π*(0.4)^{2}} = 4.5 * 10^{4} N/C </math>

===Difficult===

[[File:circle_2.png]]

A thin plastic disk located with a center at <math> <0.1, 0.25, 0> </math> m is uniformly charged with an excess of <math>7.46 * 10^{-6} </math> electrons. Find the radius of the disk if it generates an electric field of <math> E = 2.2 * 10^{3} N/C </math> at a location of <math> <0.1, -0.5, 0> </math> m?

Since we don't know the radius of the disk, we cannot be sure that using the approximation would be accurate, because we don't know if <math> z << R </math>. We must use the full formula,

<math> E = \frac{1}{2\epsilon_0}\frac{Q}{πR^2}[1 - \frac{z}{(R^2+z^2)^{1/2}}]</math>

We start by calculating Q, knowing that <math> Q = e * -1.602 * 10^{-19} = -1.2 * 10^{-6} C </math>

Next, we find z, which is essentially the distance from the center of the disk to the observation point.

<math> z = mag( <0.1, -0.5, 0> - <0.1, 0.25, 0> ) = 0.75 </math>

Finally, we use the original equation for electric field to solve for R. <math> R = 2.6 </math> m. If we solve it using the approximated formula,

<math> E = \frac{Q}{2\epsilon_0A} </math>

We calculate <math> R = 3.13 </math>, which has a significant amount of error from the original formula calculation. This indicates that using the approximation in this scenario would not have been appropriate.

==Connectedness==
Uniformly charged disks are a fundamental component to capacitors (which is two disks placed parallel to each other), which are found everywhere in physics courses as well as the real world. They are key parts of LC circuits, RC circuits, batteries, signal processing, and can even produce light.

===Charged Disks in Television===
Capacitors and charged plates used to play a heavy role in televisions! Charged plates were used to accelerate electrons, which were then manipulated and directed by magnetic fields to control what area of the television screen they would strike. As they hit the screen, a certain chemical would glow a specific color, generating an image. So every time you woke up early on saturday morning (before the advent of flat screen televisions), your episodes of Spongebob and Fairly OddParents was brought to you by capacitors!

===Charged Disks in Chemical Engineering===
The electric fields of capacitors may seem completely unrelated to chemical engineering. But actually, there are a surprising number of applications within chemical engineering. For example, much of the instrumentation used in chemical processes rely on capacitors and their ability to generate constant electric fields between two charged plates.

===Industrial Applications===
One of the primary careers in chemical engineering focuses on operating, maintaining, and optimizing a chemical plant or refinery. In a refinery, there are thousands of things you have to monitor throughout the plant. Every pipe and vessel holding liquid or gas is susceptible to corrosion, and any breach in containment could jeapordize thousands of lives. One of the most common causes of corrosion is ionic contaminants in water. Before a method was found to monitor and troubleshoot the ion concentrations in water, there were numerous dangerous corrosion events that put people's lives in danger.

Now, we monitor the conductivity of water in a refinery. This is done using CAPACITORS. Essentially, capacitors are used to monitor the changing electric field and dielectric properties of the water running through it. This data is then analyzed to find the amount of ionic contaminant in the water, allowing the chemical engineer to prevent corrosion at the source.

==History==
The first capacitor was invented in 1746. It was called the Leyden Jar, and it was essentially a glass jar wrapped in aluminum on both sides. They were used initially to store charge, almost as a battery. With the advent of radio technologies, capacitors played a key role in producing and receiving radio signals. Thus, they were heavily improved due to the demand from the radio industry.

[[File:dielectric_2.png]]

As technology progressed, capacitors were developed that were smaller and more compact, allowing broad uses and applications. These ranged from diverse circuits to incredibly powerful transmitters.

== See also ==

[[Capacitor]]

[[Charged Ring]]

===Further reading===

[https://www.webassign.net/ebooks/mi4/toc.html? Matter and Interactions 3rd Edition]

==External Links and References==

[https://phet.colorado.edu/en/simulations/category/physics PhET Simulations]

[http://www.phy.uct.ac.za/demonline/virtual/index-e.html VPython Physics Programs]

[https://en.wikipedia.org/wiki/Capacitor Further Online Reading]

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