Work done by electric field to move capacitor plate

[polls_king23]

Homework Statement

two capacitor plates separated by a distance d are moved closer to each other by a distance ε<<d due to their mutual attraction. Find the work done by the electric field in the process and also the energy lost by the field

Relevant Equations

energy U=1/2(AV^2ε0)(1/d)
where A is the area of each plate and V is the voltage

so the result I got is the energy lost by the field right??

 

[berkeman]

Welcome to PF.

When the capacitance changes, what happens to the Voltage?

Also, it's good to learn to post using LaTeX to help make your math equations easier to read. You can click the "LaTeX Guide" link below the Edit window to see how it's done. The basics are pretty easy to learn.

$$W_i = \frac{\epsilon_0 A {V_i}^2}{2d}$$

 

[berkeman]

(You can click the Reply link in my post above to see the LaTeX that I used. For stand-along lines use $$ before and after the LaTeX. For in-line math, use ## before and after.)

 

[polls_king23]

Also, it's good to learn to post using LaTeX to help make your math equations easier to read. You can click the "LaTeX Guide" link below the Edit window to see how it's done. The basics are pretty easy to learn.

yes I will, I just joined today. I will get to it as soon as possible

When the capacitance changes, what happens to the Voltage?

voltage will also change, by the relation c=q/v

 

[gneill]

The problem doesn't directly state whether or not the capacitor is isolated or connected to a constant source voltage (to maintain a constant potential difference as the plates are moved). The relevant equation intimates that the voltage is held constant, but as it is a relevant equation and not part of the problem statement itself, I think this makes the question a bit ambiguous.

If the capacitor is taken to be isolated, perhaps looking at its stored energy in terms of constant charge on the plates would be more appropriate?