Physics - Capacitor Question with Solution | TestHub

From the given circuit diagram, capacitors $C_1$, $C_2$, and $C_3$ are connected in series. This series combination is connected in parallel with capacitor $C_4$ across the battery of voltage $V$.

The equivalent capacitance $C_s$ of the series branch containing $C_1$, $C_2$, and $C_3$ is given by:

$\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

Substituting the given values:

$\frac{1}{C_s}=\frac{1}{1C}+\frac{1}{2C}+\frac{1}{3C}=\frac{6+3+2}{6C}=\frac{11}{6C}$

$C_s=\frac{6C}{11}$

Since $C_1$, $C_2$, and $C_3$ are in series, the charge on each of them is the same and is equal to the total charge on this series branch. The voltage across this branch is $V$.

Charge on $C_2$, $Q_2=C_s\times V=\frac{6CV}{11}$

Capacitor $C_4$ is connected directly across the battery, so the voltage across it is also $V$.

Charge on $C_4$, $Q_4=C_4\times V=4CV$

The ratio of charges on $C_2$ and $C_4$ is:

$\frac{Q_2}{Q_4}=\frac{\frac{6CV}{11}}{4CV}=\frac{6}{11\times 4}=\frac{6}{44}=\frac{3}{22}$

Answer: $\frac{3}{22}$