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Question

(A) Initial capacitance $C = 50\,\mu F = 50 	imes 10^{-6}\,F$ and initial potential $V = 100\;V$.Initial charge $Q = CV = 50 	imes 10^{-6} 	imes 10

Vedclass · Accepted Answer

$25 	imes 10^{-2}\,J$

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(A) Initial capacitance $C = 50\,\mu F = 50 	imes 10^{-6}\,F$ and initial potential $V = 100\;V$.Initial charge $Q = CV = 50 	imes 10^{-6} 	imes 100 = 5 	imes 10^{-3}\,C$.Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.When the distance between the plates is doubled,the new capacitance $C' = C/2 = 25\,\mu F$.The new potential $V' = Q/C' = Q/(C/2) = 2V = 200\;V$.The work done by the external agent is the change in stored energy: $W = U_f - U_i$.$U_i = \frac{1}{2}CV^2 = \frac{1}{2} 	imes 50 	imes 10^{-6} 	imes (100)^2 = 0.25\;J$.$U_f = \frac{1}{2}C'V'^2 = \frac{1}{2} 	imes (C/2) 	imes (2V)^2 = \frac{1}{2} 	imes (C/2) 	imes 4V^2 = CV^2 = 0.50\;J$.$W = 0.50 - 0.25 = 0.25\;J = 25 	imes 10^{-2}\,J$.

The plates of a parallel plate capacitor of capacity $50\,\mu F$ are charged to a potential of $100\;V$ and then disconnected from the battery. The distance between the plates is then doubled. How much work is done in doing so?

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