# Homework Help: Parallel Plate Capacitor

1. Sep 15, 2010

### ahazen

Given: There are two parallel plates. The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 × 10-3 m2, and the spacing between the plates is 6 × 10-3 m. There is no dielectric between the plates. Calculate the energy stored in the capacitor.

I found C is 2.95 x10^-12 by taking ((8.85x10^-12)(2 x10^-3))/(6x10^-3).

I know:
the electric field is uniform and that V=E*d
The potential difference is just the magnitude of the electric field times the distance you move in the electric field

When I try to solve for V or PE, I keep getting the Power of ten error.

Thank you:)

2. Sep 15, 2010

### Delphi51

The value you have for C looks good to me.
What formula are you using for the energy of the capacitor?
Note that in V = E*d, the E is not energy. Also, the potential V is not the energy; it is the energy per charge.

3. Sep 15, 2010

### ahazen

I'm not exactly sure what formula to use. I am kind of stuck.

4. Sep 15, 2010

### Delphi51

5. Sep 15, 2010

### ahazen

Thank you so much for your help:)

I figured it out. I had to find the Voltage which is 600. Then plug it into the equation: PE=1/2CV^2

6. Sep 15, 2010

### Delphi51

Most welcome!