E between parallel plates

In summary, the conversation discusses the behavior of electric fields and charge densities in a system involving two conducting plates. The first plate has a charge density of sigma 1 on each side, resulting in an electric field of sigma 1/epsilon. When a second plate is brought near, the charge on the opposite sides of each plate is attracted to the interior side, resulting in one side of each plate having a charge density of 0 and the other side having a charge density of 2*sigma 1. This causes the electric field between the plates to double to 2*sigma 1/epsilon. The conversation then raises questions about why the electric fields don't add up and why the field is 0 on the outside of the
  • #1
wumple
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I'm not sure if this qualifies as a 'homework question'. There is no specific problem...I have a question about something in the text.

It gives the situation of a conducting plate with charge density sigma 1 on each side. The E field on each side due to the conducting plate is = sigma 1/epsilon.

When a second conducting plate (plate 2) is brought near, the charge on the opposite sides of each of the plates is attracted to the interior side. So now one side of each plate has charge density 0 and one side has charge density 2 * sigma 1 (or sigma 2).

Then it says that the E field between the plates is equal to 2 *sigma 1/ epsilon or twice what it was before the second plate was brought over.

But why wouldn't the second plate generate a field with equal magnitude, making the total E field in between the plates = 4 * sigma 1 / epsilon? Why is the total E field in between the two plates only calculated from one plate, and not considering both?



Then this website: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Capacitors/ParallCap.html [Broken]

Says that the E fields are equal to (sigma 1)/2 epsilon and that the E fields add. So the book says that the charge densities double but the E fields don't add, and this website says that the charge densities stay the same but the E fields add.

Also, the E field is 0 on the outside of the plates. The website says it is because the E fields cancel. The book says it is because there is no excess charge on the outside of the plates since it all went to the inside. but why the E field be zero just because there is no charge near it? There's still charge on the inside of the plate, just a little farther away.

What's really going on? I'm confused! Help!
 
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  • #2
Consider a single conducting infinite sheet with surface charge density ##\sigma_1## on each face and total charge density ##2\sigma_1##. The electric field it generates is obtained from Gauss's law. The electric flux through the usual "pillbox" is ##\Phi_E=2EA## and the enclosed charge is ##q_{enc}=2\sigma_1A##. Application of Gauss's law gives the magnitude of the electric field, ##E=\sigma_1/\epsilon_0##. The direction of the field is perpendicular to the sheet and away from it on either side. (we assume that ##\sigma_1>0.##) We can view this as the superposition field produced by two sheets, each generating field ##E=\dfrac{\sigma_1}{2\epsilon_0}.##

So to model this, we charge two sheets, bring them close to each other and let the surface charges go wherever they want. We will consider four surfaces each producing its own field. Using subscripts 1 and 2 to identify the sheets and L and R for "left" and "right", we have $$\sigma_{1L}+\sigma_{1R}=2\sigma_1~~~~~(1);~~\sigma_{2L}+\sigma_{2R}=2\sigma_2~~~~~(2)$$Now look at the drawing below. We know that inside conductor 1 at point A the electric field must be zero. By superposition, it is also the sum of four contributions from the four distributions, one to the right from ##\sigma_{1L}## and the other three to the left.

ParallelPlates.png

We write for the total field at A, $$\frac{\sigma_{1L}}{2\epsilon_0}-\frac{\sigma_{1R}}{2\epsilon_0}-\frac{\sigma_{2L}}{2\epsilon_0}-\frac{\sigma_{2R}}{2\epsilon_0}=0$$Using equations (1) and (2) and simplifying,$$\sigma_{1L}-(2\sigma_1-\sigma_{1L})-2\sigma_2=0~\rightarrow~\sigma_{1L}=\sigma_1+\sigma_2$$Then$$\sigma_{1R}=2\sigma_1-\sigma_{1L}=\sigma_1-\sigma_2.$$ You can follow the same procedure to find the remaining two distributions by demanding that the electric field be zero at point B or you can note that there is left-to-right symmetry which means that you can swap subscripts "1" and "2" and subscripts "L" and "R" to get,
$$\sigma_{2R}=\sigma_2+\sigma_1~;~~~~~\sigma_{2L}=\sigma_2-\sigma_1.$$Once you have the four distributions, you can find the electric field anywhere. For example, between the plates,$$E_{between}=\frac{1}{2\epsilon_0}(\sigma_{1L}+\sigma_{1R}-\sigma_{2L}-\sigma_{2R})=\frac{\sigma_1-\sigma_2}{\epsilon_0}$$This method is straightforward and can be easily extended to more than two plates. If one of the distributions is given to you as negative, then just flip its sign in the above equations.

On edit 10/12/2020
Crossed out misleading statement in first sentence. The charge density is the same on each face; the total charge is ##2\sigma_1A##.
 
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1. What is E between parallel plates?

E between parallel plates refers to the electric field strength between two parallel conducting plates. It is a measure of the force per unit charge that a test charge would experience in the space between the plates.

2. How is E between parallel plates calculated?

E between parallel plates is calculated using the equation E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the distance between the plates.

3. What factors affect the value of E between parallel plates?

The value of E between parallel plates is affected by the distance between the plates, the potential difference between the plates, and the material of the plates. It is also affected by any charges present on the plates or in the surrounding environment.

4. What is the significance of E between parallel plates?

E between parallel plates is an important concept in understanding the behavior of electric fields. It is used in various applications, such as in capacitors, and is also a fundamental concept in understanding the behavior of charged particles in electric fields.

5. How is E between parallel plates related to the concept of electric flux?

E between parallel plates is directly proportional to the electric flux between the plates. Electric flux is a measure of the flow of electric field lines through a surface, and it is defined as the product of the electric field strength and the area of the surface. As the electric field strength between the plates increases, so does the electric flux.

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