# Work and Energy Loss; Electrostatics

1. May 15, 2012

### mateomy

Griffith's Problem 2.40 (a) and (b)

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance ε, as a result of their mutual attraction.

a) Use

$$P= \frac{\epsilon_0}{2}E^2$$

to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.

b) Use

$$\frac{\epsilon_0}{2}E^2 = energy per unit volume$$

to express the energy lost by the field in this process.

I solved the problems in my (they're both the same answer in the answer key) way getting

$$\frac{(q^2)\epsilon}{2A\epsilon_0}$$

But Griffiths doesn't do that, he doesnt even delve into what E is, which I solved ambiguously using A. His answer was:

$$\frac{\epsilon_0}{2}(E^2)A\epsilon$$

Is it alright that I expanded my E?

2. May 16, 2012

### rude man

Well, not really. The poblem demands that the answer be in terms of A and E.

However: was it specified whether the potential difference across the plates was kept constant, or was the voltage source used to charge the capacitor removed before the displacement took place? Makes a big difference ....

3. May 16, 2012

### mateomy

It didn't specify. What I posted was as much as he wrote in for the particular question. But the point you brought about about A and E was overlooked and that makes sense.

4. May 16, 2012

### Steely Dan

No, it makes a small difference (infinitesimally small, to be precise ;-P).

5. May 17, 2012

### rude man

Well, if you don't consider 2:1 as big, then OK, it makes no big difference.