Work and Energy Loss; Electrostatics

[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

<br /> P= \frac{\epsilon_0}{2}E^2<br />

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

b) Use

<br /> \frac{\epsilon_0}{2}E^2 = energy per unit volume<br />

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

<br /> \frac{(q^2)\epsilon}{2A\epsilon_0}<br />

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

<br /> \frac{\epsilon_0}{2}(E^2)A\epsilon<br />

Is it alright that I expanded my E?

 

[rude man]

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

<br /> P= \frac{\epsilon_0}{2}E^2<br />

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

b) Use

<br /> \frac{\epsilon_0}{2}E^2 = energy per unit volume<br />

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

<br /> \frac{(q^2)\epsilon}{2A\epsilon_0}<br />

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

<br /> \frac{\epsilon_0}{2}(E^2)A\epsilon<br />

Is it alright that I expanded my E?

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 ...

 

[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.

 

[Steely Dan]

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 ...

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

 

[rude man]

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

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