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Grade: 12
        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
2 years ago

Answers : (1)

Kushagra Tiwari
11 Points
							
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
2 years ago
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