# 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 :

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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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