Electric Field between Two Thin Conducting Plates

  • #1
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Homework Statement


Two thin conducting plates, each 25.0 cm on a side, are situated parallel to one another and 5.0 mm apart. If ##10^{-11}## electrons are moved from one plate to the other, what is the electric field between the plates?


Homework Equations


$$\vec E(P) = \frac{1}{4\pi {\varepsilon}_0}\int_{Surface} \frac{\sigma dA}{r^2}\hat r$$

The Attempt at a Solution


I thought of summing the electric field between the two plates, but I do not know the surface charge density ##(\sigma)##, and I do not know how to incorporate the ##10^{-11}## electrons into the equation.

Thank you in advance.
 
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Answers and Replies

  • #2
Since the plate separation is much smaller than the plate size, you can consider the plates as infinite. The electric field of an infinite plate can be easily found using Gauss's law, or you could just look it up.
I thought of summing the electric field between the two plates
Yes, use superposition to find the field of both plates.
how to incorporate the ##10^{-11}## electrons into the equation.
You can't have less than one electron. Do you mean ##10^{11}## electrons?
 
  • #3
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Yes, I believe it meant ##10^{11}##, but because I'm not sure, I want to calculate with using a variable n for the number of electrons and e for the charge of an electron. I understand that I can consider infinite plates, but my main problem is finding a value for sigma.

Do you have any ideas?

Thank you in advance, and thank you for noticing the error with the number of electrons.
 
  • #5
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##\sigma## would be changing right? Because depending on the number of electrons that move between the plates, the charge distribution would change. So if n electrons moved they would spread of an area A over the surface, but I don't know the area that they will be over.
 
  • #6
I meant is ##\sigma## a function of position or is it uniform across the plate?
So if n electrons moved they would spread of an area A over the surface, but I don't know the area that they will be over.
You're on the right track here. If the plates are square with side length 25.0cm, then what is the area?
 
  • #7
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Oh, I see. So ##\sigma## would be uniform across the plate, and the area should be ##(25.0 cm)^2##. If I understand, then would this be correct?

$$\vec E = \frac{1}{4\pi {\varepsilon}_0} \frac{\sigma A}{r^2} = \frac{A}{4\pi {\varepsilon}_0 r^2}( \frac{q_2}{A}-\frac{q_1}{A}) = \frac{ne}{4\pi {\varepsilon}_0 r^2}$$

r would be the distance between the plates, so 5.0 mm, ne would be the charge of n electrons. ##q_1## and ##q_2## would be the charges of the plates, and ##A=(25.0 cm)^2##

If that's right, should it be a negative or positive electric field?

Thank you
 
  • #8
Oh, I see. So ##\sigma## would be uniform across the plate, and the area should be ##(25.0 cm)^2##.
Yes
If I understand, then would this be correct?

$$\vec E = \frac{1}{4\pi {\varepsilon}_0} \frac{\sigma A}{r^2} = \frac{A}{4\pi {\varepsilon}_0 r^2}( \frac{q_2}{A}-\frac{q_1}{A}) = \frac{ne}{4\pi {\varepsilon}_0 r^2}$$
##r## is not constant in the integral, so it can't be factored out. Again, I would recommend finding ##\mathbf{E}## using Gauss's law since actually integrating over the position ##r## will be tricky.
should it be a negative or positive electric field?
It depends how the coordinate system is set up. Remember ##\mathbf{E}## points away from positive charge and towards negative charge.
 
  • #9
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rr is not constant in the integral, so it can't be factored out. Again, I would recommend finding E\mathbf{E} using Gauss's law since actually integrating over the position rr will be tricky.
If I were to use Gauss's Law, would I use the following equation?

$$\frac{Q}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA \Rightarrow \frac{ne}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA$$

Or would it be an open surface? If ##Q \neq ne##, then how would a value for Q be calculated?

Thank you.
 
  • #10
It is always a closed surface. You can use the surface of a rectangular prism or a cylinder as the Gaussian surface.
 
  • #11
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It is always a closed surface. You can use the surface of a rectangular prism or a cylinder as the Gaussian surface.
So then:

$$\frac{Q}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA \Rightarrow \frac{\sigma A}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA = A \vec E \Rightarrow \frac{\sigma}{\varepsilon_0} = \vec E = \frac{ne}{A\varepsilon_0}$$

Or does ##\sigma \neq \frac{ne}{A}##? If not, then how should I go about finding ##\sigma##?

Thank you.
 
  • #12
So then:

$$\frac{Q}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA \Rightarrow \frac{\sigma A}{\varepsilon_0}=\oint_{Surface} \vec E \dot \,dA = A \vec E \Rightarrow \frac{\sigma}{\varepsilon_0} = \vec E = \frac{ne}{A\varepsilon_0}$$

Or does ##\sigma \neq \frac{ne}{A}##? If not, then how should I go about finding ##\sigma##?

Thank you.
Remember there are two surfaces with area ##A## for the Gaussian surface, one above the plane and one below. So your answer is nearly right but off by a factor 1/2.
$$E=\frac{ne}{2A\epsilon_{0}}$$
This is the magnitude of the electric field produced by one of the planes.
 
  • #13
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Remember there are two surfaces with area ##A## for the Gaussian surface, one above the plane and one below. So your answer is nearly right but off by a factor 1/2.
$$E=\frac{ne}{2A\epsilon_{0}}$$
This is the magnitude of the electric field produced by one of the planes.
So if I want to find the electric field between both plates, would I not sum it to become ##\vec E=\frac{ne}{A\epsilon_{0}}##?
 
  • #14
So if I want to find the electric field between both plates, would I not sum it to become ##\vec E=\frac{ne}{A\epsilon_{0}}##?
For both plates, you sum it and the 1/2 factor goes away. I wasn't sure which step you were on in post #11.
 
  • #15
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For both plates, you sum it and the 1/2 factor goes away. I wasn't sure which step you were on in post #11.
Okay that makes sense, but how would I determine the direction of the electric field?
Since the charge is negative, that means the electric field is negative. If both planes are parallel to the y-axis and perpendicular to the x-axis, then does that mean the electric field is going in the ##-\hat j## direction?
 
  • #16
It depends how your diagram is laid out. One of the plates is negative but the other is positive. So if you draw a vector pointing from the positive plate to the negative plate, in which direction does it point?
 
  • #17
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Okay. Thank you so much for your help, I think I was able to get a better idea of how to solve these problems, and am able to conceptualize them better.
 

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