Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance $\epsilon $, as a result of their mutual attraction.
(a) Use Eq. 2.52 to express the work done by electrostatic forces, in terms of the field $E $, and the are of the plates, $A $.
(b) Use Eq. 2.46 to express the energy lost by the field in this process.
Remark: This problem is supposed to be easy, but it contains the embryo of an alternative derivation of Eq. 2.52, using conservation of energy.
\[P=\frac{\epsilon _0}{2}E^2 \tag{Eq. 2.52, page 101} \] \[\frac{\epsilon _0}{2}E^2 =\text{energy per unit volume} \tag{Eq. 2.46, page 94} \]This is an excellent problem to put dimensional analysis into practice!
(a) For this first part, we recall that work is force times displacement, and pressure can be used to define a force:
\[\text{Force} =\text{Pressure} \times \text{Area} \]So
\begin{align*} W &= \text{Force} \times \text{Distance} \\ &= \text{Pressure} \times \text{Area} \times \text{displacement} \\ &= \frac{\epsilon _0}{2}E^2 \times A\times \epsilon \\ &= \frac{\epsilon _0}{2}E^2A\epsilon \end{align*}(b) In a similar way, if we multiply the energy per unit volume times the change in volume, we will get the energy lost. The change in volume is
\[\Delta V=A\epsilon \]So
\begin{align*} W &= \text{Energy density} \times \text{Volume} \\ &= \frac{\epsilon _0}{2}E^2 A\epsilon \end{align*}Giving the same answer as in part (a).