Two capacitors \(\mathrm{C}_1=3 \mu \mathrm{F}\) and \(\mathrm{C}_2=2 \mu \mathrm{F}\) are connected in series across d.c. source of \(100 \mathrm{~V}\). The ratio of the potential across \(C_2\) to \(C_1\) is

\(\begin{array}{ll}
\therefore \quad & \mathrm{C}_1=3 \mu \mathrm{F} \text { and } \mathrm{C}_2=2 \mu \mathrm{F} \\
& \mathrm{C}_{\text {series }}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{6}{5} \mu \mathrm{F} \\
& \text { Also, } \mathrm{Q}=\mathrm{CV} \\
\mathrm{Q} & =\mathrm{C}_{\text {series }} \times \mathrm{V} \\
& =\frac{6}{5} \times 100=120 \mu \mathrm{C}
\end{array}\)
\(\mathrm{Q}\) will be the same across both the capacitors as they are in series.
\(\therefore \quad\) Potential across capacitors,
\(\begin{aligned}
& \mathrm{V}_1=\frac{\mathrm{Q}}{\mathrm{C}_1}=\frac{120}{3}=40 \mathrm{~V} \\
& \mathrm{~V}_2=\frac{\mathrm{Q}}{\mathrm{C}_2}=\frac{120}{2}=60 \mathrm{~V} \\
& \therefore \quad \mathrm{V}_2: \mathrm{V}_1=60: 40=3: 2
\end{aligned}\)