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(A) Let the capacitance of each identical air capacitor be $C$. Initially, both are in series with a battery of e.m.f. $V$. The equivalent capacitance

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

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(A) Let the capacitance of each identical air capacitor be $C$. Initially, both are in series with a battery of e.m.f. $V$. The equivalent capacitance is $C_{eq} = C/2$. When one capacitor is filled with a dielectric of constant $K$, its new capacitance becomes $C' = KC$. The new equivalent capacitance is $C_{eq}' = \frac{C \cdot KC}{C + KC} = \frac{KC}{K + 1}$. The total charge supplied by the battery is $Q = C_{eq}' V = \frac{KCV}{K + 1}$. Since the capacitors are in series, the same charge $Q$ flows through both. The potential difference across the air capacitor (the one not filled with dielectric) is $V_1 = \frac{Q}{C} = \frac{KCV}{C(K + 1)} = \frac{KV}{K + 1}$.

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

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