Two capacitors $A$ and $B$ each having a dielectric constant $2$ are connected in series. When they are connected across a $230 V$ D.C. supply it is found that the potential difference across $A$ is $130 V$ and that across $B$ is $100 V$. If the dielectric in the smaller capacitor is replaced by another dielectric of dielectric constant $5$, what will be the new values of the potential difference across each?

In series the charge on each capacitor is $230 V$. Therefore, potential difference given by $V=\frac{q}{C}$, the capacitors are in inverse ratio of their capacitor. Hence

$V_1=\frac{q}{C_1}=130 V$ and

$$\begin{gathered} V_2=\frac{q}{C_2}=100 V \\ \Rightarrow \frac{C_2}{C_1}=\frac{V_1}{V_2}=\frac{130}{100} \end{gathered}$$

$\Rightarrow \frac{C_1}{C_2}=\frac{10}{13}$                       ……(1)

and so the smaller capacitance is $C_1$.

As capacitance of a capacitor is proportional to the dielectric constant ($K$), so

New capacitance $C{\text{'}}_1$ of the first capacitor is

$C{\text{'}}_1=\frac{5}{2}{C}_1 \left\{\because \frac{C{\text{'}}_1}{C_1}=\frac{K{\text{'}}_1}{K_1}=\frac{5}{2}\right\}$

New capacitance of second capacitor $C{\text{'}}_2$ does not change, so

$C{\text{'}}_2=C_2$

$\Rightarrow \frac{C{\text{'}}_1}{C{\text{'}}_2}=\frac{{\displaystyle \frac{5}{2}}{C}_1}{C_2}=\frac{5}{2}\times \frac{10}{13}=\frac{25}{13}$                  {using (1)}

If $V{\text{'}}_1$ and $V{\text{'}}_2$ are the respective potential differences, then

$\frac{V{\text{'}}_1}{V{\text{'}}_2}=\frac{C{\text{'}}_2}{C{\text{'}}_1}=\frac{13}{25}$                 ……(2)

Also, $V{\text{'}}_1+V{\text{'}}_2=230 V$        ……(3)

Solving (2) and (3), we get

$V{\text{'}}_1=78.7 V$ and $V{\text{'}}_2=151.3 V$