A parallel plate capacitor of capacitance 12. | vedclass

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(C) Initial capacitance $C_0 = 12.5 \ pF$ and initial potential $V = 12.0 \ V$.Initial charge $Q = C_0 V = 12.5 	imes 10^{-12} 	imes 12 = 150 	imes

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(C) Initial capacitance $C_0 = 12.5 \ pF$ and initial potential $V = 12.0 \ V$.Initial charge $Q = C_0 V = 12.5 	imes 10^{-12} 	imes 12 = 150 	imes 10^{-12} \ C$.Initial energy $U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} 	imes 12.5 	imes 10^{-12} 	imes (12)^2 = 900 	imes 10^{-12} \ J$.After disconnecting the battery,the charge $Q$ remains constant. When the dielectric slab is inserted,the new capacitance becomes $C_f = \epsilon_r C_0 = 6 	imes 12.5 \ pF = 75 \ pF$.The final energy $U_f = \frac{Q^2}{2 C_f} = \frac{Q^2}{2 \epsilon_r C_0} = \frac{U_i}{\epsilon_r} = \frac{900 	imes 10^{-12}}{6} = 150 	imes 10^{-12} \ J$.The change in potential energy is $\Delta U = U_i - U_f = 900 	imes 10^{-12} - 150 	imes 10^{-12} = 750 	imes 10^{-12} \ J$.

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