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(A) Initially,two capacitors of capacity $C$ are in series. The equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} =

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$\frac{C}{2}\left[\frac{K-1}{K+1}\right]$

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(A) Initially,two capacitors of capacity $C$ are in series. The equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_1 = \frac{C}{2}$.After filling one capacitor with a dielectric of constant $K$,its new capacity becomes $KC$. The new equivalent capacitance $C_2$ is given by $\frac{1}{C_2} = \frac{1}{C} + \frac{1}{KC} = \frac{1}{C} \left(1 + \frac{1}{K}\right) = \frac{K+1}{KC}$.Thus,$C_2 = \frac{KC}{K+1}$.The change in effective capacitance $\Delta C = C_2 - C_1 = \frac{KC}{K+1} - \frac{C}{2}$.$\Delta C = C \left[ \frac{K}{K+1} - \frac{1}{2} \right] = C \left[ \frac{2K - (K+1)}{2(K+1)} \right] = \frac{C}{2} \left[ \frac{K-1}{K+1} \right]$.

Two parallel plate air capacitors of the same capacity $C$ are connected in series to a battery of emf $E$. Then one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacitance of the series combination is

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