Capacitors

JEE Main level, single correct, medium

173 of 180

Question 173 of 180

A parallel plate capacitor having capacitance 12pF12 pF is charged by a battery to a potential difference of 10V10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.56.5 is slipped between the plates. The work done by the capacitor on the slab is

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Answer: (C) 508pJ508pJ

Initial energy of capacitor
Ui=12q2c{U}_{i}=\dfrac{1}{2}\dfrac{{q}^{2}}{c}
=12×120×12012=600pJ=\dfrac{1}{2}\times \dfrac{120\times 120}{12}=600pJ
Since battery is disconnected so charge remain same. Final energy of capacitor
Uf=12q2ck{U}_{f}=\dfrac{1}{2}\dfrac{{q}^{2}}{ck}
=12×120×12012×6.5=92pJ=\dfrac{1}{2}\times \dfrac{120\times 120}{12\times 6.5}=92pJ
W+Uf=UiW+{U}_{f}={U}_{i}
W=508pJW=508pJ

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