Question

A parallel plate capacitor having capacitance $12\, pF$ is charged by a battery to a potential difference of $10\, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates the work done by the capacitor on the slab is :

• A. 692 pJ
• B. 60 pJ
• C. 508 pJ
• D. 560 pJ

Answer 1

Answer: (C)

Intial energy of capacitor
$U_{i} = \frac{1}{2} \frac{v^{2}}{c} $
$ = \frac{1}{2} \times \frac{120 \times120}{12} =600J $
Since battery is disconnected so charge remain same.
Final energy of capacitor
$ U_{f} = \frac{1}{2} \frac{v^{2}}{c} $
$ = \frac{1}{2} \times\frac{120 \times120}{12\times6.5} = 92 $
$ W + U_{f} =U_{i} $
$ W =508\, J $