Fernando Garcia

PHYS405 - Electricity & Magnetism I

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 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.

\begin{matrix} (Eq. 2.52, page 101) & P = \frac{ϵ_{0}}{2} E^{2} \end{matrix}

\begin{matrix} (Eq. 2.46, page 94) & \frac{ϵ_{0}}{2} E^{2} = energy per unit volume \end{matrix}

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:

Force = Pressure \times Area

\begin{aligned} W & = Force \times Distance \\ = Pressure \times Area \times displacement \\ = \frac{ϵ_{0}}{2} E^{2} \times A \times ϵ \\ = \frac{ϵ_{0}}{2} E^{2} A ϵ \end{aligned}

(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

Δ V = A ϵ

\begin{aligned} W & = Energy density \times Volume \\ = \frac{ϵ_{0}}{2} E^{2} A ϵ \end{aligned}

Giving the same answer as in part (a).