Three identical parallel plate air capacitors are connected in series to a battery of e.m.f. V. If two capacitors are completely filled with dielectric material of constant k, then potential difference across the third capacitor will become

Let the capacitance of each identical air capacitor be $C$.

When two capacitors are completely filled with a dielectric material of constant $k$, their capacitances become $kC$ and $kC$. The third capacitor remains an air capacitor with capacitance $C$.

The equivalent capacitance $C_{eq}$ of the three capacitors connected in series is given by:

$\dfrac{1}{C_{eq}} = \dfrac{1}{kC} + \dfrac{1}{kC} + \dfrac{1}{C}$

$\dfrac{1}{C_{eq}} = \dfrac{2}{kC} + \dfrac{k}{kC} = \dfrac{k+2}{kC}$

$C_{eq} = \dfrac{kC}{k+2}$

The total charge $Q$ drawn from the battery of e.m.f. $V$ is:

$Q = C_{eq}V = \dfrac{kCV}{k+2}$

In a series circuit, the charge on each capacitor is the same. Therefore, the potential difference $V_3$ across the third capacitor (capacitance $C$) is:

$V_3 = \dfrac{Q}{C} = \dfrac{\dfrac{kCV}{k+2}}{C} = \dfrac{kV}{k+2}$

Answer: $\dfrac{kV}{(k+2)}$