Physics - Capacitor Question with Solution | TestHub

From the given circuit diagram, the capacitor $C_3$ is connected directly across the battery $V$.

Therefore, the potential difference across $C_3$ is $V$.

The charge on $C_3$ is given by:
$Q_3=C_{3}V=3CV$

Capacitors $C_1$ and $C_2$ are connected in series, and this series combination is connected in parallel with $C_3$ across the battery $V$.

The equivalent capacitance of the series combination of $C_1$ and $C_2$ is:
$C_{12}=\frac{C_1{C}_2}{C_1+C_2}=\frac{1C\times 2C}{1C+2C}=\frac{2C^2}{3C}=\frac{2}{3}C$

The potential difference across this series combination is also $V$. In a series circuit, the charge on each capacitor is the same and equal to the total charge of the combination.

The charge on $C_1$ is:
$Q_1=C_{12}V=\frac{2}{3}CV$

The ratio of the charge on capacitor $C_3$ to that on $C_1$ is:
$\frac{Q_3}{Q_1}=\frac{3CV}{\frac{2}{3}CV}=\frac{3\times 3}{2}=\frac{9}{2}=4.5$

Answer: 4.5