Electric Field Between two Parallel Conducting Plates of Equal Charge

[rtareen]

Attached is the subsection of the book I am referring to. The previous section states that the electric field magnitude at any point set up by a charged nonconducting infinite sheet (with uniform charge distribution) is $E = \frac{\sigma}{2\epsilon_0}$.

Then we move onto the attached section, conducting parallel plates of equal and opposite charge.

We know that the charge density for each face of either plate when isolated is $\sigma_1$. When we bring them near each other the charges all move to the interior faces so that each interior face has charge density $2\sigma_1 = \sigma$.

The book doesn't explain, but I believe they then use the formula for a nonconducting sheet to to find the field between them at any point. Since the fields both point in the same direction that's the same as saying the charge density is $4\sigma_1 = 2\sigma$. Is that right?

Next, by plugging this into the field formula for a nonconducting sheet, which we can do because the two faces act as two nonconductors as the charges on the faces of the opposite plates will not move away from each other, we get that:

$E = \cfrac{4\sigma_1}{2\epsilon_0} = \cfrac{2\sigma_1}{\epsilon_0} = \cfrac{\sigma}{\epsilon_0}$

Let me know if I understood this correctly.

 

[Charles Link]

I think you have most everything right other than the charge density is not $4 \sigma_1$. The charge density is $+2 \sigma_1$ on one plate, and $-2 \sigma_1$ on the other. You can use the non-conducting sheet formula along with superposition to get the $E =4 \sigma_1/(2 \epsilon_o)$ between the plates.

 

[rtareen]

I think you have most everything right other than the charge density is not $4 \sigma_1$. The charge density is $+2 \sigma_1$ on one plate, and $-2 \sigma_1$ on the other. You can use the non-conducting sheet formula along with superposition to get the $E =4 \sigma_1/(2 \epsilon_o)$ between the plates.

Thank you!