# Electric Potential, Work, and Capacitance

1. Feb 10, 2008

### foxjwill

1. The problem statement, all variables and given/known data
A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.60 µC.

How much work is required to quadruple the separation?

2. Relevant equations

$$C = \frac{Q}{V}$$

$$V = Ed$$

3. The attempt at a solution

In previous parts of the question, I was able to find, correctly, the potential difference before ($$V_i = 2.83\times 10^3\ \textsc{V}$$) and after ($$V_f = 1.13 \times 10^4\ \textsc{V}$$).

What's confusing me is what equation to use for work.

I don't think I'm supposed to use $$W = \frac{1}{2}CV^2$$ because that finds the work needed to charge the capacitor.

The only other equation I can think of is $$W = -\frac{QV}{d}$$, but I'm not sure how to implement it. I've tried changing it to $$W = -QE$$, but beyond that, I'm stuck.

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