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(D) Let the capacitances be $C_1 = x, C_2 = 2x, C_3 = 3x, C_4 = 4x$. From the circuit,$C_3$ and $C_2$ are in series,and this combination is in series

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$3: 22$

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(D) Let the capacitances be $C_1 = x, C_2 = 2x, C_3 = 3x, C_4 = 4x$. From the circuit,$C_3$ and $C_2$ are in series,and this combination is in series with $C_1$. Let the equivalent capacitance of the upper branch be $C_{up}$. $\frac{1}{C_{up}} = \frac{1}{C_3} + \frac{1}{C_2} + \frac{1}{C_1} = \frac{1}{3x} + \frac{1}{2x} + \frac{1}{x} = \frac{2+3+6}{6x} = \frac{11}{6x}$. So,$C_{up} = \frac{6x}{11}$. The charge on the upper branch is $Q_{up} = C_{up} V = \frac{6xV}{11}$. Since $C_2$ is in the upper branch,the charge on $C_2$ is $Q_2 = Q_{up} = \frac{6xV}{11}$. The capacitor $C_4$ is connected directly across the voltage source $V$,so the charge on $C_4$ is $Q_4 = C_4 V = 4xV$. The ratio of charges is $\frac{Q_2}{Q_4} = \frac{6xV/11}{4xV} = \frac{6}{11 	imes 4} = \frac{6}{44} = \frac{3}{22}$.

Four capacitors are connected as shown in the figure. If $C_1, C_2, C_3$ and $C_4$ are in the ratio of $1: 2: 3: 4$,then the ratio of the charges on the capacitors $C_2$ and $C_4$ is

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