Physics Book - User contributions [en]

Charged Disk

2017-11-29T19:38:08Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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:Oomo.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

2017-11-29T19:35:59Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Ima.PNG]]

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:Oomo.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?]]

File:Oomo.png

2017-11-29T19:34:25Z

Kfarley32: Kfarley32 uploaded a new version of "File:Oomo.png"

Charged Disk

2017-11-29T19:32:34Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Ima.PNG]]

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:Oomo.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]]

===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?]]

File:Oomo.png

2017-11-29T19:31:46Z

Kfarley32:

Charged Disk

2017-11-29T19:30:36Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Ima.PNG]]

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.

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]]

===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

2017-11-29T19:30:10Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Ima.png]]

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.

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]]

===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

2017-11-29T19:29:33Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Ima.jpg]]

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.

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]]

===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

2017-11-29T19:28:55Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

[[File:Why.jpg]]

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.

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]]

===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?]]

File:Ima.PNG

2017-11-29T19:28:26Z

Kfarley32:

Charged Disk

2017-11-29T19:19:16Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

[[File:Why.jpg]]

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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?]]

File:Why.jpg

2017-11-29T19:18:40Z

Kfarley32:

Charged Disk

2017-11-29T19:13:57Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

[[File:Neww.jpg]]

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.

===A Medium Example Problem===

[[File:Yee.jpg]]

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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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?]]

File:Yee.jpg

2017-11-29T19:12:57Z

Kfarley32:

File:Neww.jpg

2017-11-29T19:07:07Z

Kfarley32: Kfarley32 uploaded a new version of "File:Neww.jpg"

Charged Disk

2017-11-29T19:05:25Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?
[[File:Neww.jpg]]
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.

===A Medium Example Problem===

A thin plastic disk 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> <0.04, 0.06, 0> </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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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?]]

File:Neww.jpg

2017-11-29T19:05:10Z

Kfarley32:

Charged Disk

2017-11-29T19:00:05Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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:Ringss.jpg]]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

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.

===A Medium Example Problem===

A thin plastic disk 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> <0.04, 0.06, 0> </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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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?]]

File:Ringss.jpg

2017-11-29T18:59:39Z

Kfarley32:

Charged Disk

2017-11-29T18:48:30Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpg]]

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===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

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.

===A Medium Example Problem===

A thin plastic disk 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> <0.04, 0.06, 0> </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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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

2017-11-29T18:48:11Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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:Ddisk.jpeg]]

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===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

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.

===A Medium Example Problem===

A thin plastic disk 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> <0.04, 0.06, 0> </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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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?]]

File:Ddisk.jpg

2017-11-29T18:47:37Z

Kfarley32:

Charged Disk

2017-11-29T18:24:26Z

Kfarley32:

'''Edited Fall 2017 by Kathryn Farley'''

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==

In this section, we will do a step-by-step process of calculating the electric field of a uniformly charged 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:Disk.gif]]

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===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

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==

===An Easy Conceptual Question===
In what circumstance would you not use the approximate formula for calculating the electric field and have to use another formula?

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.

===A Medium Example Problem===

A thin plastic disk 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> <0.04, 0.06, 0> </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>

===A Difficult Example Problem===

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.

===1. How is this topic connected to something you are interested in?===
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!

===2. How is this connected to your major?===
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.

===3. Is there an interesting industrial application?===
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.

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]]

===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 Rod

2017-11-29T18:23:16Z

Kfarley32:

'''Claimed by )'''

Changes: Reformatting to match the template, adding a computational model, adding examples, and some grammatical/spelling fixes. Figures used in the examples were drawn by me.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

==The Main Idea: Electric Field of Distributed Charges==

===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

Below, the process to find the electric field of a uniformly charged thin rod is carried out.

====The Algorithm====

The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

===Special Case: A Very Long Rod===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The method of calculating the electric field of a uniformly charged rod is most accurate and necessary for problems with rods that aren't very long relative to the observation distance. For problems where the rod is long enough, the electric field for the rod can be approximated as <math> E=\frac{1}{4\pi\epsilon_0} \cdot \frac{2Q/L}{r}</math>.

===Special Case: Uniform Thin Rod At An Arbitrary Location===
When finding the electric field at an arbitrary location (as in not on the midline as shown above) due to a uniformly charged thin rod, we follow the same procedure that was used in the earlier example. The only difference is found at the observation location, which will have an x- and y-component instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.
You can see how when mathematically finding the electric field around a rod, you treat the rod as a series of point charges. This is how it was done in the explanations above, and you can see the difficult example below to put this idea into practice.

==Examples==

===Simple===

This problem is meant to give you practice using the pre-derived formulas. While this is good for understanding the differences between the use of the actual formula and the approximation, you will almost always have to derive it yourself on the exam for a different scenario.

'''Problem Statement'''

A thin plastic rod of length 2.2 m is rubbed all over with wool, and acquires a charge of 89 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 15 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.

'''Solution'''

'''(a) Exact formula'''

From the exam formula sheet: <math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r \sqrt{r^2 + (L/2)^2}} </math> (<math> r </math> perpendicular from the center)

<math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{89e^{-9}}{.15 \sqrt{.15^2 + (2.2/2)^2}} </math>

<math> |\vec{E_{rod}}| = 4810.03 N/C </math>

'''(b) Approximate Formula'''

From the exam formula sheet: <math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2Q/L}{r} </math> (if <math> r << L) </math>

<math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2(89e^{-9}/2.2)}{.15} </math>

<math> |\vec{E_{rod}}| = 4854.55 N/C </math>

This shows that while the approximation formula can be accurate for rod lengths much greater than the distance from the rod, in this case, the distances are too close for it to be a good approximation.

===Middling===

This example is meant to walk you through the steps of setting up an integration problem.

'''Problem Statement'''

A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of <math> -Q </math> is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <math> ‹ 0, y, 0 › </math> due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram.

[[File:Middling_.jpg]]

Use the following as necessary: <math> x, y, dx, A, Q </math>. Remember that the rod has charge <math>-Q</math>.

(a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?

(b) What is the amount of charge <math>dQ</math> on the small piece of length <math> dx </math>?

(c) What is the vector from source to observation location?

(d) What is the distance from the source to the observation location?

(e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

'''Solution'''

'''Part A'''

<math> \lambda = - \frac{Q}{2A} </math>

'''Part B'''

<math> dQ = - \frac{Q}{2A} dx </math>

'''Part C'''

<math> \vec{r} = < -x, y, 0> </math>

'''Part D'''

<math> d = \sqrt{(-x)^2 + y^2} </math>

'''Part E'''

<math> x </math>

===Difficult===

This problem is meant to combine the first two examples to give you practice doing an actual exam problem, where you're presented with an unfamiliar setup, and have to derive the formula for electric field yourself.

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem_update.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

===Charged Rod and Pith Ball===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==See Also==

Along with calculating electric fields of uniformly charged thin rods, many real-world examples require the electric fields of other shapes and objects to be known. Information on this wiki about some of these topics can be found below.

[[Point Charge]]

[[Electric Dipole]]

[[Charged Ring]]

[[Charged Spherical Shell]]

[[Charged Disk]]

[[Charged Capacitor]]

==History and Applications==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law." A portrait of Coulomb is shown below.

[[File:coloumb.jpg]]

The model of a uniformly charged rod can be applicable to situations where the rod is interacting with other charged objects, whether they are large or microscopic, such as charged particles, like protons or electrons, travelling in the rod's field.

==References==

'''Text Sources'''

Chabay, Ruth and Sherwood, Bruce. ''Matter and Interactions, Volume II.'' 4th ed. New Jersey: Wiley, 2015. Print.

Plaza.obu.edu. N. p., 2016. Web. 4 Nov. 2016. <http://plaza.obu.edu/corneliusk/up2/effcr.pdf>

Phys.uri.edu. N. p., 2016. Web. 4 Nov. 2016. <http://www.phys.uri.edu/gerhard/PHY204/tsl31.pdf>

Stmary.ws. N. p., 2016. Web. 4 Nov. 2016. <http://stmary.ws/HighSchool/Physics/home/notes/electricity/staticElectricity/induction.htm>

'''Image Sources'''

N.a. "Electric field of a charged rod at many locations." n.d. Shodor Education Foundation, n.a. Shodor Visioning WORKshop, Computational Labs for Electric and Magnetic Interactions (M&I Vol. 2). Web. 16 Apr. 2016. [http://compphysed.shodor.org/documents/Info%20materials/WrkGrp-1/VPython_EM/index.html Electric field of a charged rod at many locations]

N.a. "Coulomb's Law." n.d. Georgia State University, Atlanta. HyperPhysics. Web. 17 Apr. 2016. [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html Coulomb's Law]

25: Video Tutor: Charged Rod and Aluminum Can "25: Video Tutor: Charged Rod And Aluminum Can". Masteringmasteringphysics.blogspot.com. N. p., 2014. Web. 4 Nov. 2016.

Work in progress

--[[User:Spennell3|Spennell3]] ([[User talk:Spennell3|talk]]) 14:47, 2 December 2015 (EST)

--[[User:Acowart6|Acowart6]] ([[User talk:Acowart6|talk]]) 20:22, 17 April 2016 (EDT)

--[[User:Emoauro3|Emoauro3]] ([[User talk:Emoauro3|talk]]) 22:22, 2 November 2016 (EDT)

Charged Rod

2017-11-29T18:20:43Z

Kfarley32:

'''Claimed by Kathryn Farley (Fall 2017)'''

Changes: Reformatting to match the template, adding a computational model, adding examples, and some grammatical/spelling fixes. Figures used in the examples were drawn by me.

Some objects, such as rods, can be modeled as a uniformly charged object in order to calculate the electric field at some observation location. The following wiki page provides an overview of electric fields created by uniformly charged thin rods, briefly presenting its inception, some specific cases, and proof of concept experiments. The case of a uniformly charged thin rod is a fundamental example of electric field patterns and calculations within physics. Its implications can be applied to other charged objects, such as rings, disks, and spheres.

==The Main Idea: Electric Field of Distributed Charges==

===A Uniformly Charged Thin Rod===

In practical situations, objects have charges spread all over their surface. When going about calculating the electric fields of these objects, we can either use one of two processes: numerical summation or integration, or dividing an object into many pieces and summing the individual pieces' electric field contributions. As with point charges, the direction of the field is determined by the sign of the object's charge (positive-points away, negative-points toward) and the size of the field is determined by the observation distance and the magnitude of the object's charge.

The process of finding the electric field due to charge distributed over an object has four steps:

1. Divide the charged object into small pieces. Make a diagram and draw the electric field <math>\Delta \vec{E}</math> contributed by one of the pieces.

2. Choose an origin and the axes. Write an algebraic expression for the electric field <math>\Delta \vec{E}</math> due to one piece.

3. Add up the contributions of all pieces, either numerically or symbolically.

4. Check that the result is physically correct.

[[File:2D_Charged_Rod.png]]
[[File:3D_Charged_Rod.png]]

The two figures above depict both a two-dimensional and three-dimensional view of the electric field of a uniformly positively charged rod. The red figure represents a rod and the green arrows represent the electric field direction and magnitude.

Below, the process to find the electric field of a uniformly charged thin rod is carried out.

====The Algorithm====

The process of calculating a uniformly charged rod's electric field is tedious, but breaking the process down into several steps makes the task at hand easier. Consider a uniformly charged thin rod of length <math>L</math> and positive charge <math>Q</math> centered on and lying along the x-axis. The rod is being observed from above at a point on the y-axis.

'''Step 1: Divide the Distribution into Pieces; Draw <math>\Delta \vec{E}</math>'''

Imagine dividing the rod into a series of very thin slices, each with the same charge <math>\Delta Q</math>. This charge <math>\Delta Q</math> is a small part of the overall charge. Picture it as a point charge. Each slice contributes its own electric field, <math>\Delta E</math>. Summing all these individual slices of <math>E</math> gives you the total electric field of the rod. This process is the same as taking an integral, as each thickness approaches 0 and the the number of slices approaches infinity. Note that in this example, the variable that is changing for each slice is its x-coordinate.

'''Step 2: Write an Expression for the Electric Field Due to One Piece'''

The second step is to write a mathematical expression for the field <math>\Delta E</math> contributed by a single slice of the rod. We use the formula of the electric field for a point charge because we are imagining each slice as a point charge. First, determine <math>r</math>, the vector pointing from the source to the observation location. For our example, this is <math> r = obs - source = <0,y,0> - < x,0,0> = <-x,y,0></math>. Now use this to calculate the magnitude and direction of <math>r</math>. So <math>|\vec{r}| = \sqrt{(-x)^2 + y^2} = \sqrt{x^2 + y^2}</math> and <math>\hat{r} = \frac{\vec{r}}{\hat{r}} = \frac{< -x,y,0>}{\sqrt{x^2 + y^2}} </math>. <math> \hat{r}</math> is the vector portion of the expression for the field. The scalar portion is <math> \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{|\vec{r}|^2}</math>. Thus the expression for one slice of the rod is:
<math> \Delta \vec{E} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\Delta Q}{(\sqrt{x^2+y^2})^{3/2}} \cdot < -x,y,0> </math>.

'''Determining <math>\Delta Q</math> and the integration variable'''

In the first step, we determined that the changing variable for this rod was its x-coordinate. This means the integration variable is <math> dx</math>. We need to put this integration variable into our expression for the electric field. More specifically, we need to express <math>\Delta Q</math> in terms of the integration variable. Recall that the rod is uniformly charged, so the charge on any single slice of it is: <math>
\Delta Q = (\frac{\Delta x}{L})\cdot Q</math>. This quantity can also be expressed in terms of the charge density.

'''Expression for <math> \Delta \vec{E}</math>

Substitute the expression for the integration variable into the formula for the electric field of one slice. Separating the equation into separate x and y components, we get <math> \Delta \vec{E_x} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{-x}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math> and <math> \Delta \vec{E_y} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx </math>. Note that we have replaced <math> \Delta x </math> with <math> dx</math> in preparation for integration.

'''Step 3: Add Up the Contributions of All the Pieces'''

The third step is to sum all of our slices. We can go about this in two ways. One way is with numerical summation, or separating the object into a finite number of small pieces, calculating the individual contributing electric fields, and then summing them. Another, more precise method is to integrate. Most of the work of finding the field of a uniformly charged object is setting up this integral. If you have reached the correct expression to integrate, the rest is simple math. The bounds for integration are the coordinates of the start and stop of the rod. In this example the bounds are from <math>-L/2</math> to <math>+L/2</math>. So the expression is <math> \int\limits_{-L/2}^{L/2}\ \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{L} \cdot \frac{y}{(\sqrt{x^2+y^2})^{3/2}} \cdot dx. </math> Solving this gives the final expression <math> E_y = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x} \cdot \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. Note that the field parallel to the x axis is zero. This can be observed due to the symmetry of the problem.
This equation can be written more generally as <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r} \cdot \frac{1}{(\sqrt{r^2+ (L/2)^2})} </math> where r represents the distance from the rod to the observation location.

'''Step 4:Checking the Result'''

Finally, the fourth step is to check the result. The units should be the same as the units of the expression for the electric field for a single point particle <math> E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2} </math>.
Our answer has the right units, since <math> \frac{1}{(\sqrt{x^2+ (L/2)^2})} </math>. has the same units of <math>\frac{Q}{r^2}</math>
Is the direction qualitatively correct? We have the electric field pointing straight away from the midpoint of the rod, which is correct, given the symmetry of the situation. The vertical component of the electric field should indeed be zero.

===Special Case: A Very Long Rod===

For special cases, an approximated formula to find the electric field will be provided. The steps to derive this formula are the same as those outlined above. The method of calculating the electric field of a uniformly charged rod is most accurate and necessary for problems with rods that aren't very long relative to the observation distance. For problems where the rod is long enough, the electric field for the rod can be approximated as <math> E=\frac{1}{4\pi\epsilon_0} \cdot \frac{2Q/L}{r}</math>.

===Special Case: Uniform Thin Rod At An Arbitrary Location===
When finding the electric field at an arbitrary location (as in not on the midline as shown above) due to a uniformly charged thin rod, we follow the same procedure that was used in the earlier example. The only difference is found at the observation location, which will have an x- and y-component instead of only an x-component as in the example above.

=== A Computational Model===
[https://trinket.io/embed/glowscript/c59cde2427 This simulation] depicts the electric field around a charged rod.
You can see how when mathematically finding the electric field around a rod, you treat the rod as a series of point charges. This is how it was done in the explanations above, and you can see the difficult example below to put this idea into practice.

==Examples==

===Simple===

This problem is meant to give you practice using the pre-derived formulas. While this is good for understanding the differences between the use of the actual formula and the approximation, you will almost always have to derive it yourself on the exam for a different scenario.

'''Problem Statement'''

A thin plastic rod of length 2.2 m is rubbed all over with wool, and acquires a charge of 89 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 15 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.

'''Solution'''

'''(a) Exact formula'''

From the exam formula sheet: <math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r \sqrt{r^2 + (L/2)^2}} </math> (<math> r </math> perpendicular from the center)

<math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{89e^{-9}}{.15 \sqrt{.15^2 + (2.2/2)^2}} </math>

<math> |\vec{E_{rod}}| = 4810.03 N/C </math>

'''(b) Approximate Formula'''

From the exam formula sheet: <math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2Q/L}{r} </math> (if <math> r << L) </math>

<math> |\vec{E_{rod}}| = \frac{1}{4\pi\epsilon_{0}} \frac{2(89e^{-9}/2.2)}{.15} </math>

<math> |\vec{E_{rod}}| = 4854.55 N/C </math>

This shows that while the approximation formula can be accurate for rod lengths much greater than the distance from the rod, in this case, the distances are too close for it to be a good approximation.

===Middling===

This example is meant to walk you through the steps of setting up an integration problem.

'''Problem Statement'''

A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of <math> -Q </math> is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <math> ‹ 0, y, 0 › </math> due to the rod. Following the procedure discussed in the textbook, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram.

[[File:Middling_.jpg]]

Use the following as necessary: <math> x, y, dx, A, Q </math>. Remember that the rod has charge <math>-Q</math>.

(a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?

(b) What is the amount of charge <math>dQ</math> on the small piece of length <math> dx </math>?

(c) What is the vector from source to observation location?

(d) What is the distance from the source to the observation location?

(e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

'''Solution'''

'''Part A'''

<math> \lambda = - \frac{Q}{2A} </math>

'''Part B'''

<math> dQ = - \frac{Q}{2A} dx </math>

'''Part C'''

<math> \vec{r} = < -x, y, 0> </math>

'''Part D'''

<math> d = \sqrt{(-x)^2 + y^2} </math>

'''Part E'''

<math> x </math>

===Difficult===

This problem is meant to combine the first two examples to give you practice doing an actual exam problem, where you're presented with an unfamiliar setup, and have to derive the formula for electric field yourself.

'''Problem Statement:'''

A very thin plastic rod of length <math>L</math> is rubbed with cloth and becomes uniformly charged with a total charge <math>+Q</math>.

[[File:Hard_problem_update.jpg]]

(a) Consider an arbitrary piece of rod of length <math>dx</math> located at a position x on the rod. Determine the electric field <math>\vec{E}</math> from this piece at observation location "*", a distance <math>w</math> to the right of the end of the rod and on the x-axis as indicated in the diagram.

(b) Write down and solve the integral to determine the net electric field <math>\vec{E}</math> of the rod at location "*".

'''Solution'''

'''Part A:'''

<math>\vec{r} = \vec{r_*}-\vec{r_{dx}} = < L+w, 0, 0 > - < x, 0, 0 > = < L+w-x, 0, 0 > = (L+w-x)\hat{x}</math>

<math> |\vec{r}| = L+w-x </math> ; <math>\hat{r} = < 1, 0, 0 > = \hat{x} </math>

<math>\lambda = \frac{charge}{length} = \frac{Q}{L} = \frac{dQ}{dx} </math>

<math> dQ = \frac{Q}{L} dx </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{dQ}{|\vec{r}|^2} \hat{r} = \frac{1}{4\pi\epsilon_{0}} \frac{(Q/L)dx}{(L+w-x)^2} \hat{x} </math>

<math> d\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x} </math>

'''Part B:'''

<math> \int d\vec{E} = \int_{0}^{L}\frac{1}{4\pi\epsilon_{0}} \frac{Qdx}{L(L+w-x)^2}\hat{x}
= \frac{Q}{4\pi\epsilon_{0}L} \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x}</math>

To solve the integral, use the following (found on your test formula sheet): <math>\int \frac{dx}{(x+a)^2} = \frac{-1}{a+x} </math>

<math> \int_{0}^{L} \frac{dx}{(L+w-x)^2}\hat{x} = \frac{1}{L+w-x} (x_1 = 0, x_2 = L) =
\frac{1}{L+w-L} - \frac{1}{L+w} = \frac{1}{w} - \frac{1}{L+w} = \frac {L+w-w}{w(L+w)} = \frac{L}{w(L+w)} </math>

<math> \vec{E} = \frac{Q}{4\pi\epsilon_{0}L} \frac{L}{w(L+w)} \hat{x} </math>

<math> \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{w(L+w)} \hat{x} </math>

==Connectedness: Practical Experiments==
The following two DIY experiments are shown to better understand and visualize the physical consequences of electric fields.

===Charged Rod and Aluminum Can===

In our first example we set up an experiment using two charged rods placed to the left and right of an aluminum can, distanced by a length d. If one of the rods is positively charged and the other is negatively charged, what will the can do? Because the positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can, and the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod, the net force on the can is zero. Thus the can will stay still. The setup is depicted in the image below.

[[File:Can_and_Two_Charged_Rods_(1).png]]

Next, lets consider a setup but with both rods having equal positive charge, as shown in the image below. What will the can do in this situation?

[[File:Can_and_Two_Charged_Rods_(2).png]]

Again the can will also stay still, but this time it is because the polarization force between two objects is always attractive.

So in what scenario will the can move, and what time of movement will the can exhibit? Considering the first setup, imagine this time we initially touch the negatively charged rod and the can for a brief moment. Holding the rods at equal distance on either side of the can, the can will now roll toward the positively charged rod. This is because the can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod.

===Charged Rod and Pith Ball===

Another DIY experiment to visualize the effects of an electric field are shown in the video embedded below, which depicts the interaction between an initially neutral pith ball hanging on a string from a stand, and a charged rod.
[http://www.youtube.com/watch?v=aeiqw81kGio Interaction between a Charged Rod and Pith Ball]

==See Also==

Along with calculating electric fields of uniformly charged thin rods, many real-world examples require the electric fields of other shapes and objects to be known. Information on this wiki about some of these topics can be found below.

[[Point Charge]]

[[Electric Dipole]]

[[Charged Ring]]

[[Charged Spherical Shell]]

[[Charged Disk]]

[[Charged Capacitor]]

==History and Applications==

Physicists and scientists make use of electric fields and charged objects all the time. Many times, we may need to know which objects are contributing how much charge in certain areas. Charged objects may attract or repel (depending on the signs of their charge), so we often need to know how objects will interact with each other based on their charges. The phenomenon of this interaction, or electric force between charged particles, was finally confirmed and stated as a law in 1785 by French physicist Charles-Augustin de Coulomb, hence "Coulomb's Law." A portrait of Coulomb is shown below.

[[File:coloumb.jpg]]

The model of a uniformly charged rod can be applicable to situations where the rod is interacting with other charged objects, whether they are large or microscopic, such as charged particles, like protons or electrons, travelling in the rod's field.

==References==

'''Text Sources'''

Chabay, Ruth and Sherwood, Bruce. ''Matter and Interactions, Volume II.'' 4th ed. New Jersey: Wiley, 2015. Print.

Plaza.obu.edu. N. p., 2016. Web. 4 Nov. 2016. <http://plaza.obu.edu/corneliusk/up2/effcr.pdf>

Phys.uri.edu. N. p., 2016. Web. 4 Nov. 2016. <http://www.phys.uri.edu/gerhard/PHY204/tsl31.pdf>

Stmary.ws. N. p., 2016. Web. 4 Nov. 2016. <http://stmary.ws/HighSchool/Physics/home/notes/electricity/staticElectricity/induction.htm>

'''Image Sources'''

N.a. "Electric field of a charged rod at many locations." n.d. Shodor Education Foundation, n.a. Shodor Visioning WORKshop, Computational Labs for Electric and Magnetic Interactions (M&I Vol. 2). Web. 16 Apr. 2016. [http://compphysed.shodor.org/documents/Info%20materials/WrkGrp-1/VPython_EM/index.html Electric field of a charged rod at many locations]

N.a. "Coulomb's Law." n.d. Georgia State University, Atlanta. HyperPhysics. Web. 17 Apr. 2016. [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html Coulomb's Law]

25: Video Tutor: Charged Rod and Aluminum Can "25: Video Tutor: Charged Rod And Aluminum Can". Masteringmasteringphysics.blogspot.com. N. p., 2014. Web. 4 Nov. 2016.

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