Three identical parallel plate air capacitors are connected in...

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