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A parallel plate capacitor of plate area A and plate separation d is charged to 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

Vinayak Saxena , 7 Years ago
Grade 12
anser 1 Answers
Kushagra Tiwari

Last Activity: 7 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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