Electric Potential, Work, and Capacitance

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Homework Statement


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?


Homework Equations



[tex]C = \frac{Q}{V}[/tex]

[tex]V = Ed[/tex]


The Attempt at a Solution



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

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

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

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

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