Electric field due to charges between 2 parallel infinite planes

[Tesla In Person]

TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point

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Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A.
I know that the field depends only on z since there is translational invariance in x and y directions because the planes are infinite.
View attachment 365748
So applying gauss law : Q_enc is the volume of the cube of side a and face area A. That gives Q_enc= p*A*a and left hand side of gauss's law gives 2*A*E. I excluded the areas facing y and x directions since the field is pointing in the z direction so only the 2 faces facing in that direction are cut be the field lines. I get 2AE = p*A*a / epsilon and simplifying this gives E = p*a / 2*epsilon. I'm not sure if this is correct. If it's incorrect, I want to know where I made mistakes. I'm doing this for the first time and I'm struggling with these question especially when I have to find the right gaussian surface for the problem.

 

[berkeman]

Your attachments did not seem to work. Can you try again? What file type are they?

 

[Tesla In Person]

Your attachments did not seem to work. Can you try again? What file type are they?

I don't know why they are not opening , Hope its fixed now. Its png files.

 

[kuruman]

. . . and left hand side of gauss's law gives 2*A*E.

There are three regions of space:

1. Inside the distribution $~-\frac{a}{2} \leq z \leq\frac{a}{2}$
2. To the left of the distribution: $~z\leq-\frac{a}{2}$
3. To the right of the distribution: $~z\geq\frac{a}{2}$.

You need to find the electric field in each region. This means writing the electric field vector $\mathbf E(z)$ as a function of $z$.
Hint: Is there a value of $z$ at which the electric field is zero?

 

[Tesla In Person]

There are three regions of space:

1. Inside the distribution $~-\frac{a}{2} \leq z \leq\frac{a}{2}$
2. To the left of the distribution: $~z\leq-\frac{a}{2}$
3. To the right of the distribution: $~z\geq\frac{a}{2}$.

You need to find the electric field in each region. This means writing the electric field vector $\mathbf E(z)$ as a function of $z$.
Hint: Is there a value of $z$ at which the electric field is zero?

Is it zero at the origin , z=0? But why would it be 0 at the origin , is it just a convention or is the strength of the electric field really 0 there ? The LaTeX doesn't appear in the message so I don't understand the points 1. 2. and 3.

 

[kuruman]

Is it zero at the origin , z=0? But why would it be 0 at the origin , is it just a convention or is the strength of the electric field really 0 there ? The LaTeX doesn't appear in the message so I don't understand the points 1. 2. and 3.

For some reason the LaTeX works in "Preview" mode only. Here it is

Answer the following four questions and you will see why the electric field is zero at z = 0.

1. In what direction does the electric field point in region 2, to the left of the distribution?
2. In what direction does the electric field point in region 3, to the right of the distribution?
3. In what direction does the electric field point inside the slab, to the left of the origin?
4. In what direction does the electric field point inside the slab, to the right of the origin?