Two identical parallel plate capacitors of ca | vedclass

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(B) Initially,the equivalent capacitance is $C_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.The charge delivered by the battery is $Q_1 = C_{eq} E =

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$\frac{k-1}{2(k+1)} \cdot C E$

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(B) Initially,the equivalent capacitance is $C_{eq} = \frac{C \cdot C}{C + C} = \frac{C}{2}$.The charge delivered by the battery is $Q_1 = C_{eq} E = \frac{C E}{2}$.When one of the capacitors is filled with a dielectric of constant $k$,its new capacitance becomes $kC$. The new equivalent capacitance of the series combination is:$C'_{eq} = \frac{C \cdot (kC)}{C + kC} = \left( \frac{k}{k+1} \right) C$.Since the battery remains connected,the new charge on the combination is:$Q_2 = C'_{eq} E = \left( \frac{k}{k+1} \right) C E$.The amount of charge that flows through the battery is the change in charge:$\Delta Q = Q_2 - Q_1 = \left( \frac{k}{k+1} - \frac{1}{2} \right) C E$.$\Delta Q = \left( \frac{2k - (k+1)}{2(k+1)} \right) C E = \frac{k-1}{2(k+1)} C E$.

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