# Parallel-plate capacitor

1. Sep 11, 2011

### ajmCane22

1. The problem statement, all variables and given/known data

A parallel-plate capacitor has plates with an area of 383 cm2 and an air-filled gap between the plates that is 1.51 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.
(a) How much energy is stored in the capacitor?

(b) The separation between the plates is now increased to 4.53 mm. How much energy is stored in the capacitor now?

(c) How much work is required to increase the separation of the plates from 1.51 mm to 4.53 mm?

3. The attempt at a solution

This is what I did:

C = εA/d to find C and then used E = 1/2CV^2 to find E. Then for part (c) I subtracted the value for part (b) from the value for part (a). I got the following values:
(a)3.712e-5 J
(b)1.236e-5 J
(c)2.476e-5 J

(b) and (c) are incorrect and I'm not sure what I'm doing wrong. (a) and (b) should be the same equation, correct? I don't understand why I'm getting part (a) right and part (b) wrong. Please help.

2. Sep 11, 2011

### kuruman

When the plates are separated by some additional distance, which of the quantities Q, C, V remain the same and which ones change?

3. Sep 11, 2011

### ajmCane22

Oh! The voltage changes, so I need to calculate the second voltage, right?

4. Sep 11, 2011

### kuruman

Yes, the voltage changes but the charge remains the same.