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 :


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 $

You must select option to get answer and solution

Questions from JEE Main 2019

Physics Most Viewed Questions

6. The spectrum of an oil flame is an example for ...........

KCET 2010 Dual Nature Of Radiation And Matter