Obtain the equivalent capacitance of the network in the figure. For a $300 \; V$ supply,determine the charge and voltage across each capacitor.

(N/A) Given: $C_{1} = 100 \; pF$,$C_{2} = 200 \; pF$,$C_{3} = 200 \; pF$,$C_{4} = 100 \; pF$,and supply voltage $V = 300 \; V$.
$1$. Capacitors $C_{2}$ and $C_{3}$ are in series. Their equivalent capacitance $C'$ is given by:
$\frac{1}{C'} = \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} = \frac{1}{100} \implies C' = 100 \; pF$.
$2$. $C_{1}$ and $C'$ are in parallel. Their equivalent capacitance $C''$ is:
$C'' = C_{1} + C' = 100 + 100 = 200 \; pF$.
$3$. $C''$ and $C_{4}$ are in series. The total equivalent capacitance $C_{eq}$ is:
$\frac{1}{C_{eq}} = \frac{1}{C''} + \frac{1}{C_{4}} = \frac{1}{200} + \frac{1}{100} = \frac{3}{200} \implies C_{eq} = \frac{200}{3} \; pF$.
$4$. Charge on $C_{4}$ is $Q_{4} = C_{eq} \times V = \frac{200}{3} \times 10^{-12} \times 300 = 2 \times 10^{-8} \; C$.
Voltage across $C_{4}$ is $V_{4} = \frac{Q_{4}}{C_{4}} = \frac{2 \times 10^{-8}}{100 \times 10^{-12}} = 200 \; V$.
$5$. Voltage across the parallel combination ($C_{1}$ and $C'$) is $V'' = V - V_{4} = 300 - 200 = 100 \; V$.
Thus,$V_{1} = 100 \; V$ and $V' = 100 \; V$.
Charge on $C_{1}$ is $Q_{1} = C_{1} \times V_{1} = 100 \times 10^{-12} \times 100 = 10^{-8} \; C$.
$6$. Since $C_{2}$ and $C_{3}$ are in series and have equal capacitance,the voltage $V'$ divides equally: $V_{2} = V_{3} = \frac{100}{2} = 50 \; V$.
Charge on $C_{2}$ is $Q_{2} = C_{2} \times V_{2} = 200 \times 10^{-12} \times 50 = 10^{-8} \; C$.
Charge on $C_{3}$ is $Q_{3} = C_{3} \times V_{3} = 200 \times 10^{-12} \times 50 = 10^{-8} \; C$.