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(B) Let the capacitance of each identical capacitor be $C$. When connected in series to a voltage source $V_0$,the equivalent capacitance is $C_{eq} =

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$\left( \frac{K}{K+1} \right) V_0$

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(B) Let the capacitance of each identical capacitor be $C$. When connected in series to a voltage source $V_0$,the equivalent capacitance is $C_{eq} = C/2$.Initially,the potential difference across each capacitor is $V_1 = V_2 = V_0/2$.When one capacitor is immersed in a dielectric of constant $K$,its new capacitance becomes $C' = KC$.The other capacitor remains $C$. The capacitors are still in series.The new potential difference $V_{other}$ across the capacitor with capacitance $C$ is given by the voltage divider rule:$V_{other} = V_0 \left( \frac{C'}{C + C'} \right) = V_0 \left( \frac{KC}{C + KC} \right)$$V_{other} = V_0 \left( \frac{KC}{C(1 + K)} \right) = \left( \frac{K}{K+1} \right) V_0$.

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0 \, V$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$,the potential difference between the plates of the other capacitor will change to:

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