Dielectrics in Capacitors: Difference between revisions
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This page is about dielectrics placed between two capacitor plates, and its effect on the capacitor and the circuit it's a part of. Claimed by Alex Lim (alim33) | |||
This page is about dielectrics placed between two capacitor plates, and its effect on the capacitor and the circuit it's a part of. | |||
==The Main Idea== | ==The Main Idea== | ||
| Line 7: | Line 6: | ||
===A Mathematical Model=== | ===A Mathematical Model=== | ||
C(with dielectric) = kC | |||
k = dielectric constant (since a dielectric always increases capacitance, k > 1) | |||
<math> C=\frac{Q}{V} </math> | |||
C = capacitance, Q = the charge on the capacitor, V = voltage across the plates | |||
Another way to write capacitance, | |||
<math> C = \frac{k*\epsilon_0*A}{d} </math> | |||
C = capacitance, k = dielectric constant, <math> \epsilon_0 = 8.85e-12 \frac{C^2}{N}*m^2 </math> | |||
<math> U = \frac{1}{2}CV^2 </math> | |||
U = energy | |||
==Examples== | ==Examples== | ||
A dielectric filled parallel-plate capacitor has a plate area, A = 30<math>cm^2</math>, plate separation d = 0.02 m, and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 6.00 V. Throughout the problem, use <math>\epsilon_0 = 8.85e-12 \frac{C^2}{N}*m^2 </math>. | |||
=== | A) Find the energy <math>U_1</math> of the dielectric-filled capacitor. Express your answer numerically in joules. | ||
=== | |||
=== | <math> C = \frac{k*\epsilon_0*A}{d} = \frac{3(*8.85e-12)(30/10000)}{0.02} = 3.9825e-12 </math> F | ||
<math>U_1 = = \frac{1}{2}CV^2 = 7.168e-11 </math> J | |||
B) The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy <math>U_2</math> of the capacitor at the moment when the capacitor is half-filled with the dielectric. Express your answer numerically in joules. | |||
<math> C_2 = \frac{k*\epsilon_0*A}{d} + \frac{\epsilon_0*A}{d} = \frac{1}{2}(1+k)\frac{epsilon_0*A}{d} = 2.655e-12 </math> F | |||
<math>U_2 = = \frac{1}{2}CV^2 = 4.779e-11 </math> J | |||
C) The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, <math>U_3</math>. Express your answer numerically in joules. | |||
<math> Q = C_2V = 1.593e-11</math> C | |||
<math> C_3 = \frac{\epsilon_0*A}{d} = 1.3275e-12 </math> F | |||
<math> U_3 = \frac{1}{2}\frac{Q^2}{C_3} = 9.558e-11 </math> J | |||
D) In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work, W, is ond eby the external agent acting on the dielectric? Express your answer numerically in joules. | |||
Work = W = <math>U_3</math> - <math>U_2</math> = 4.779e-11 J | |||
==Connectedness== | ==Connectedness== | ||
How is this topic connected to something that you are interested in? | |||
Capacitors are used in the design of batteries, and batteries power almost all of my electronics and devices. | |||
How is it connected to your major? | |||
In supply chain, one of the main areas of work is in factories. A lot of the machinery that would be used in the factory is powered by batteries. | |||
Is there an interesting industrial application? | |||
When batteries are being changed, the system can still have a temporary charge because a capacitor keeps the initial charge that the battery had before it dissipates over time. | |||
== | == See also == | ||
[[Capacitors]] | |||
===External links=== | ===External links=== | ||
| Line 40: | Line 82: | ||
==References== | ==References== | ||
Matter and Interactions 4th Edition | |||
http://physics.info/dielectrics/ | |||
http://physics.info/dielectrics/problems.shtml | |||
https://www.youtube.com/watch?v=rkntp3_cZl4 | |||
Latest revision as of 00:55, 6 December 2015
This page is about dielectrics placed between two capacitor plates, and its effect on the capacitor and the circuit it's a part of. Claimed by Alex Lim (alim33)
The Main Idea
What are dielectrics? Dielectrics are non-conducting material (insulator). One of the functions of a dielectric in a capacitor is the effect it has on the capacitance, a capacitor's storage potential which is measured in units called farads(F). A dielectric ALWAYS increases the capacitance, this'll be further explained below. Another function of a dielectric is the separation of the two capacitor plates because if the two plates are in contact then it would complete the circuit and no charge would be ever be stored in the capacitor plates.
A Mathematical Model
C(with dielectric) = kC
k = dielectric constant (since a dielectric always increases capacitance, k > 1)
[math]\displaystyle{ C=\frac{Q}{V} }[/math]
C = capacitance, Q = the charge on the capacitor, V = voltage across the plates
Another way to write capacitance,
[math]\displaystyle{ C = \frac{k*\epsilon_0*A}{d} }[/math]
C = capacitance, k = dielectric constant, [math]\displaystyle{ \epsilon_0 = 8.85e-12 \frac{C^2}{N}*m^2 }[/math]
[math]\displaystyle{ U = \frac{1}{2}CV^2 }[/math]
U = energy
Examples
A dielectric filled parallel-plate capacitor has a plate area, A = 30[math]\displaystyle{ cm^2 }[/math], plate separation d = 0.02 m, and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 6.00 V. Throughout the problem, use [math]\displaystyle{ \epsilon_0 = 8.85e-12 \frac{C^2}{N}*m^2 }[/math].
A) Find the energy [math]\displaystyle{ U_1 }[/math] of the dielectric-filled capacitor. Express your answer numerically in joules.
[math]\displaystyle{ C = \frac{k*\epsilon_0*A}{d} = \frac{3(*8.85e-12)(30/10000)}{0.02} = 3.9825e-12 }[/math] F
[math]\displaystyle{ U_1 = = \frac{1}{2}CV^2 = 7.168e-11 }[/math] J
B) The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy [math]\displaystyle{ U_2 }[/math] of the capacitor at the moment when the capacitor is half-filled with the dielectric. Express your answer numerically in joules.
[math]\displaystyle{ C_2 = \frac{k*\epsilon_0*A}{d} + \frac{\epsilon_0*A}{d} = \frac{1}{2}(1+k)\frac{epsilon_0*A}{d} = 2.655e-12 }[/math] F
[math]\displaystyle{ U_2 = = \frac{1}{2}CV^2 = 4.779e-11 }[/math] J
C) The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, [math]\displaystyle{ U_3 }[/math]. Express your answer numerically in joules.
[math]\displaystyle{ Q = C_2V = 1.593e-11 }[/math] C
[math]\displaystyle{ C_3 = \frac{\epsilon_0*A}{d} = 1.3275e-12 }[/math] F
[math]\displaystyle{ U_3 = \frac{1}{2}\frac{Q^2}{C_3} = 9.558e-11 }[/math] J
D) In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work, W, is ond eby the external agent acting on the dielectric? Express your answer numerically in joules.
Work = W = [math]\displaystyle{ U_3 }[/math] - [math]\displaystyle{ U_2 }[/math] = 4.779e-11 J
Connectedness
How is this topic connected to something that you are interested in?
Capacitors are used in the design of batteries, and batteries power almost all of my electronics and devices.
How is it connected to your major?
In supply chain, one of the main areas of work is in factories. A lot of the machinery that would be used in the factory is powered by batteries.
Is there an interesting industrial application?
When batteries are being changed, the system can still have a temporary charge because a capacitor keeps the initial charge that the battery had before it dissipates over time.
See also
External links
References
Matter and Interactions 4th Edition
http://physics.info/dielectrics/