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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 whole space between the plates. Find the work done on the system in the process of inserting the slab.

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 whole space between the plates. Find the work done on the system in the process of inserting the slab.

Grade:11

1 Answers

Arun
25763 Points
3 years ago
 
Initial energy – final energy = work done
And charge remains conserved = CV = A*epsilon*V/D
Therefore, W = ColdV2/2 – Q2/2Cnew= A*epsilon*V2/2D – A*epsilon*V2/2KD

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