Question
a parallel plate capacitor of plate A and plate seperation d is charged to a potential dif. V and then the battery is disconnected .a slab of dielectric constant K is then inserted between the plates of capacitor so as tofill the space bet. The plates .find the work done in this process.Joshi sir comment
initial energy of capacitor = 1/2 CV2
final energy of capacitor = 1/2 KC (V/K)2
difference ?
Read 1 Solution.
WORK DONE = ZERO BECAUSE U initial & final is same that is q2/ 2KC . WHEN THE SLAB IS REMOVED THEN ALSO THE SAME U. THUS W.d BY SYSTEM IS ZERO.
SARIKA SHARMA 12 year ago
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