A parallel plate capacitor of capacity 100 mu | vedclass

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Question

(B) Initial capacitance $C = 100 \mu F = 100 	imes 10^{-6} 	ext{ F}$ and potential $V = 50 	ext{ V}$.Initial energy stored $E_1 = \frac{1}{2} C V^2

Vedclass · Accepted Answer

$125 	imes 10^{-3}$

Vedclass · Answer

(B) Initial capacitance $C = 100 \mu F = 100 	imes 10^{-6} 	ext{ F}$ and potential $V = 50 	ext{ V}$.Initial energy stored $E_1 = \frac{1}{2} C V^2 = \frac{1}{2} 	imes 100 	imes 10^{-6} 	imes (50)^2 = 50 	imes 10^{-6} 	imes 2500 = 0.125 	ext{ J} = 125 	imes 10^{-3} 	ext{ J}$.When the distance $d$ is doubled,the new capacitance $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} = 50 \mu F$.Since the battery remains connected,the potential $V$ remains $50 	ext{ V}$.Final energy stored $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} 	imes 50 	imes 10^{-6} 	imes (50)^2 = 25 	imes 10^{-6} 	imes 2500 = 0.0625 	ext{ J} = 62.5 	imes 10^{-3} 	ext{ J}$.Charge on the capacitor initially $Q_1 = CV = 100 	imes 10^{-6} 	imes 50 = 5 	imes 10^{-3} 	ext{ C}$.Charge on the capacitor finally $Q_2 = C'V = 50 	imes 10^{-6} 	imes 50 = 2.5 	imes 10^{-3} 	ext{ C}$.Change in charge $\Delta Q = Q_2 - Q_1 = -2.5 	imes 10^{-3} 	ext{ C}$.Work done by the battery $W = \Delta Q 	imes V = (-2.5 	imes 10^{-3}) 	imes 50 = -125 	imes 10^{-3} 	ext{ J}$.The negative sign indicates that energy is returned to the battery. The magnitude of energy change is $125 	imes 10^{-3} 	ext{ J}$.

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