Q. 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 :

Solution:

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 $

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