Physics - Capacitor Question with Solution | TestHub

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:

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

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

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

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

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

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:

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

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