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Question

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Given radius $r = 10\, cm = 0.1\, m$, so $A = \pi r^2 = \pi

Vedclass · Accepted Answer

Loss of $12.5\, 	ext{ergs}$

Vedclass · Answer

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.Given radius $r = 10\, cm = 0.1\, m$, so $A = \pi r^2 = \pi(0.1)^2 = 0.01\pi\, m^2$.Initial distance $d_1 = 1\, mm = 10^{-3}\, m$. Initial capacitance $C_1 = \frac{\epsilon_0 A}{d_1}$.Final distance $d_2 = 1\, cm = 10^{-2}\, m = 10d_1$. Final capacitance $C_2 = \frac{\epsilon_0 A}{d_2} = \frac{C_1}{10}$.Since the battery remains connected, the potential $V = 100\, V$ is constant.Initial energy $U_1 = \frac{1}{2} C_1 V^2$.Final energy $U_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} (\frac{C_1}{10}) V^2 = \frac{U_1}{10}$.Change in energy $\Delta U = U_2 - U_1 = \frac{1}{2} V^2 (C_2 - C_1) = \frac{1}{2} (100)^2 (\frac{C_1}{10} - C_1) = \frac{1}{2} (10000) (-0.9 C_1) = -4500 C_1$.Using $\epsilon_0 \approx 8.854 	imes 10^{-12}\, F/m$, $C_1 = \frac{8.854 	imes 10^{-12} 	imes 0.01\pi}{10^{-3}} \approx 2.78 	imes 10^{-10}\, F$.$\Delta U = -4500 	imes 2.78 	imes 10^{-10} \approx -1.25 	imes 10^{-6}\, J$.Since $1\, J = 10^7\, 	ext{ergs}$, $\Delta U = -1.25 	imes 10^{-6} 	imes 10^7 = -12.5\, 	ext{ergs}$.The negative sign indicates a loss of energy.

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