Question

A parallel-plate capacitor whose capacitance is C = 13.5pF has a potential difference of V = 12.5Vbetween its plates. The charging battery is now disconnected and a porcelain slab with dielectricconstant κ = 6.50 is slipped between the plates. What is the potential energy of the device before andafter the slab is introduced?

Answer 1

C=13.5pF

V=12.5V

k=6.50

$U=\frac{1}{2}{CV^2}$

$$U_i=\frac{1}{2}\times13.5\times(12.5)^2pJ=1054pJ$$

When battery connect

$U=\frac{1}{2}\frac{CV^2}{K}$

$U_f=\frac{1}{2}\times\frac{13.5\times(12.5)^2}{6.50}=162pJ$

$$W=U_i-U_f=1054-162=892pJ$$