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(A) Let the initial capacitance of each capacitor be $C$. When connected in series to a battery of emf $V$,the charge $q$ on both capacitors is the sa

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$\frac{K V}{K+1}$

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(A) Let the initial capacitance of each capacitor be $C$. When connected in series to a battery of emf $V$,the charge $q$ on both capacitors is the same.Let $V_1$ be the potential difference across the air capacitor (capacitance $C$) and $V_2$ be the potential difference across the dielectric-filled capacitor (capacitance $KC$).Since they are in series,$V_1 + V_2 = V$.Using $q = CV$,we have $V_1 = \frac{q}{C}$ and $V_2 = \frac{q}{KC}$.Substituting these into the sum equation: $\frac{q}{C} + \frac{q}{KC} = V$.$\frac{q}{C} (1 + \frac{1}{K}) = V \Rightarrow \frac{q}{C} (\frac{K+1}{K}) = V$.Therefore,the charge $q = \frac{CKV}{K+1}$.The potential difference across the air capacitor is $V_1 = \frac{q}{C} = \frac{KV}{K+1}$.

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

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