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Question

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

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Solution

Initial capacitance, C=0 Ad
The energy of the capacitor before the insertion of the dielectric is given by
E1=12 CV2=120AdV2

After inserting the dielectric slab, the capacitance becomes
C1=KC=K0 Ad
and the final voltage becomes
V1=VK

Thus, the final energy stored in the capacitor is given by

E2=12K0 Ad×VK2E2=120 AV2Kd

Now,

Work done = Change in energy
∴ W = E2 - E1
W=120 AV2Kd-120 AV2dW=120 AV2d 1K-1

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