A parallel plate capacitor of capacity 100 mu | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{125}{2} 	imes 10^{-3}$

Vedclass · Answer

(A) Initial capacitance $C = 100 \mu F = 100 	imes 10^{-6} 	ext{ F}$ and voltage $V = 50 	ext{ V}$.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 $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2} = 50 \mu F = 50 	imes 10^{-6} 	ext{ F}$.Since the battery remains connected,the voltage $V$ remains constant at $50 	ext{ V}$.Final energy stored $E_2 = \frac{1}{2} C' V^2 = \frac{1}{2} 	imes 50 	imes 10^{-6} 	imes (50)^2 = 25 	imes 10^{-6} 	imes 2500 = 0.0625 	ext{ J} = 62.5 	imes 10^{-3} 	ext{ J}$.The charge on the capacitor changes from $Q_1 = CV$ to $Q_2 = C'V$. The change in charge is $\Delta Q = (C' - C)V = (50 - 100) 	imes 10^{-6} 	imes 50 = -50 	imes 10^{-6} 	imes 50 = -2500 	imes 10^{-6} 	ext{ C}$.The work done by the battery is $W = \Delta Q 	imes V = (-2500 	imes 10^{-6}) 	imes 50 = -125000 	imes 10^{-6} = -0.125 	ext{ J} = -125 	imes 10^{-3} 	ext{ J}$.The negative sign indicates that energy is returned to the battery. The magnitude of energy change supplied by the battery is $125 	imes 10^{-3} 	ext{ J}$ (returned). However,the question asks for the additional energy supplied. Since the energy of the capacitor decreases,the battery absorbs energy. The change in energy of the capacitor is $E_2 - E_1 = 62.5 	imes 10^{-3} - 125 	imes 10^{-3} = -62.5 	imes 10^{-3} 	ext{ J}$. The battery supplies $-62.5 	imes 10^{-3} 	ext{ J}$,which is $\frac{125}{2} 	imes 10^{-3} 	ext{ J}$ removed.

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