Problem 2

A parallel-plate capacitor of capacitance

C_0

(air-filled) is charged to voltage

U_0

then disconnected from the source. A dielectric slab of constant

\kappa

is then inserted, filling the gap. Find the capacitor's energy before and after insertion, and explain the change. (a) Find plate charge, capacitance after insertion, and final voltage. (b) Calculate initial and final field energies and their ratio in terms of

\kappa

. (c) Explain the mechanical work associated with the energy decrease as the slab is drawn in; distinguish field work from external work and state what is conserved after disconnection.

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

E_f=Q^2/(2\kappa C_0)=E_i/\kappa

problems.proof.analysis

The final charge remains $Q$ , while capacitance is $\kappa C_0$ , hence $E_f=E_i/\kappa$ . Keep charge fixed here; a fixed-voltage comparison would be wrong.