# Capacitance and electric fields

1. May 6, 2008

### Niles

[SOLVED] Capacitance and electric fields

1. The problem statement, all variables and given/known data
I want to find the capacitance of a parallel-plate capacitor consisting of two metal surfaces og area A held a distance d apart.

First I find the electric field, which I know is $$\frac{\sigma }{{2\varepsilon }}$$. Then I use that the potential is the line integral from minus-charge plate to positive-charge plate of electric field, so

$$V = \frac{\sigma }{{2\varepsilon }}d$$

But in my book they do not divide by 2. This is because they have used superposition, which is OK - I agree with that.

Next, we consider the following:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html#c2

Here they have used the electric field for |one| cylinder, they have not used superposition. This I do not agree with - we are still between the cylinders, so why do we not multiply by 2 here?

Last edited: May 6, 2008
2. May 6, 2008

### Niles

Oh, right.. because the charge inside our gaussian surface is the charge coming from the "inside-cylinder"...

I get it now. Thanks

3. May 6, 2008

### Cipher_101

Actually the electric field is not between two cylinders, but it is in between a cylinder and a pipe, so it is not the same case as two parallel capacitors.