- #1
mateomy
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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
[tex]
P= \frac{\epsilon_0}{2}E^2
[/tex]
to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.
b) Use
[tex]
\frac{\epsilon_0}{2}E^2 = energy per unit volume
[/tex]
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
[tex]
\frac{(q^2)\epsilon}{2A\epsilon_0}
[/tex]
But Griffiths doesn't do that, he doesn't even delve into what E is, which I solved ambiguously using A. His answer was:
[tex]
\frac{\epsilon_0}{2}(E^2)A\epsilon
[/tex]
Is it alright that I expanded my E?
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
[tex]
P= \frac{\epsilon_0}{2}E^2
[/tex]
to express the amount of work done by the electrostatic forces, in terms of E and the area of the plates, A.
b) Use
[tex]
\frac{\epsilon_0}{2}E^2 = energy per unit volume
[/tex]
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
[tex]
\frac{(q^2)\epsilon}{2A\epsilon_0}
[/tex]
But Griffiths doesn't do that, he doesn't even delve into what E is, which I solved ambiguously using A. His answer was:
[tex]
\frac{\epsilon_0}{2}(E^2)A\epsilon
[/tex]
Is it alright that I expanded my E?