# Charge density of capacitor plates

1. Dec 23, 2017

### Physgeek64

1. The problem statement, all variables and given/known data
A parallel plate capacitor is made of two flat horizontal conducting plates in a vacuum, each of area A, separated by a small gap. One plate carries a total charge 2Q, the other a total charge −Q. Find the surface charge densities on the four horizontal metal surfaces in terms of Q and A. Find the electric field between the plates, and the electric field immediately above and below the capacitor.

2. Relevant equations

3. The attempt at a solution
I would have though $\sigma_1 =\sigma_2=\frac{Q}{A}$ and $\sigma_3=\sigma_4=-\frac{Q}{2A}$ but the answer is $sigma_1=\frac{Q}{2A}$ $\sigma_2=\frac{3Q}{2A}$ $\sigma_3=\frac{-3Q}{2A}$ $\sigma_4=\frac{-Q}{2A}$ but i cannot see how they have got this

2. Dec 23, 2017

### rude man

3 equations, 3 unknowns, since the inner surface charges will obviously be equal and opposite.
2 equations should be obvious.
The 3rd equation hint: force the E field in either plate = 0.

Last edited: Dec 23, 2017
3. Dec 23, 2017

### Physgeek64

Why should the two inner plates be equal and opposite?

4. Dec 23, 2017

### rude man

BTW the answer is wrong. If σ1 + σ2 = 2Q then σ3 + σ4 = -Q, not -2Q.
(I left out A in all the σ terms).

5. Dec 23, 2017

### rude man

Because the E field is constant between the plates. Use Gauss to verify.

6. Dec 23, 2017

### Physgeek64

No, that's right. There's 2Q on one plate and -Q on the other plate

7. Dec 23, 2017

### Physgeek64

The E field would still be constant if $\sigma_2$ is not equal to $\sigma_3$ ?

8. Dec 23, 2017

### rude man

9. Dec 23, 2017

### rude man

No. Put a gaussian surface from the inside of one plate to any point between the plates. The E field is the same everywhere inside the plates.
Another argument would be that the divergence of D (and thus E) must be zero in the absence of charge.

10. Dec 23, 2017

### Physgeek64

Sorry, have I? There is 2Q on one plate and -Q on the other

11. Dec 23, 2017

### Physgeek64

Sorry if I am being really stupid but having $\sigma_2$ not equal to $sigma_3$ gives a uniform field in between the plates of strength $\frac{\sigma_2}{2\epsilon_0 } + \frac{\sigma_3}{2\epsilon_0}$ using gauss' law

12. Dec 23, 2017

### rude man

Maybe we're having problems defining which surface is which. You did not show a drawing or verbally describe which is which.
Which is why i said that if 1 and 2 are on the plate with charge 2Q then 3 and 4 must add up to -Q but your answer adds to -2Q.

Last edited: Dec 23, 2017
13. Dec 23, 2017

### Physgeek64

Ahh I see the confusion now. I think then maybe $\sigma_4=\frac{Q}{2A}$

I have define plates 1234 to be the sequence of planes met if we travel from the top of the capacitor to the bottom where the top capacitor plate has charge 2Q and the bottom charge -Q.

14. Dec 23, 2017

### rude man

I think so too. Then their answer would be correct.
PS I woud call the two plates "plates" but their surfaces "surfaces".

15. Dec 23, 2017

### rude man

I think so too. Then their answer would be correct.
PS I woud call the two plates "plates" but their surfaces "surfaces".
EDIT: sorry, I didn't see you use the word "planes" I thought it was "plates". Perfectly OK to say "planes" but I still like "surfaces". We speak of "surface charge" not "plane charge".

16. Dec 27, 2017

### Physgeek64

I'll use surfaces from now on :) I still don't see why you wouldn't have a uniform field with different charges on the inner surfaces

17. Dec 27, 2017

### rude man

Put a Gaussian cylinder between the plates. The contained charge is zero. If the E field were not uniform you would get a net non-zero flux emanating from the gaussian surfaces parallel to the plates but that is a violation of Gauss's theorem.