Obtain the equivalent capacitance of the network shown in the figure. For a 300 V supply, detere the charge and voltage across each capacitor.

Capacitance of capacitor $C_1$ is 100 pF.

Capacitance of capacitor $C_2$ is 200 pF.

Capacitance of capacitor $C_3$ is 200 pF.

Capacitance of capacitor $C_4$ is 100 pF.

Supply potential, $V=300V$

Capacitors $C_2$ and $C_3$ are connected in series. Let their equivalent capacitance be $C^{\prime }$.

$\therefore \frac{1}{C^{\prime }}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}$

$C^{\prime }=100pF$

Capacitors $C_1$ and $C^{\prime }$ are in parallel. Let their equivalent capacitance be $C^{\prime \prime }$.

$\therefore C^{\prime \prime }=C^{\prime }+C_1$

          $=100+100=200pF$

$C^{\prime \prime }$ and $C_4$ are connected in series. Let their equivalent capacitance be C.

$\therefore \frac{1}{C}=\frac{1}{C^{\prime \prime }}+\frac{1}{C_4}$

         $=\frac{1}{200}+\frac{1}{100}=\frac{2+1}{200}$

$C=\frac{200}{3}pF$

Hence, the equivalent capacitance of the circuit is $\frac{200}{3}pF$. 

Potential difference across $C^{\prime \prime }=V^{\prime \prime }$

Potential difference across $C_4=V_4$

$\therefore V^{\prime \prime }+V_4=V=300V$

Charge on $C_4$ is given by,

$Q_4=CV$

      $=\frac{200}{3}\times 10^{-12}\times 300$

      $=2\times 10^{-8}C$

      $V_4=\frac{Q_4}{C_4}$

      $=\frac{2\times 10^{-8}}{100\times 10^{-12}}=200V$

$\therefore$ Voltage across $C_1$ is given by,

$V_1=V-V_4$

     $=300-200=100V$

Hence, potential difference, $V_1$, across $C_1$ is 100 V.

Charge on $C_1$ is given by,

$Q_1=C_1{V}_1$

      $=100\times 10^{-12}\times 100$

      $=10^{-8}C$

$C_2$ and $C_3$ having same capacitances have a potential difference of 100 V together. Since $C_2$ and $C_3$ are in series, the potential difference across $C_2$ and $C_3$ is given by,

$V_2=V_3=50V$

Therefore, charge on $C_2$ is given by,

$Q_2=C_2{V}_2$

     $=200\times 10^{-12}\times 50$

     $=10^{-8}C$

And charge on $C_3$ ­is given by,

$Q_3=C_3{V}_3$

      $=200\times 10^{-12}\times 50$

      $=10^{-8}C$

Hence, the equivalent capacitance of the given circuit is $\frac{200}{3}$pF with,

$\begin{array}{cc}{Q}_1=10^{-8}C,& V_1=100V\\ Q_2=10^{-8}C,& V_2=50V\\ Q_3=10^{-8}C,& V_3=50V\\ Q_4=2\times 10^{-8}C,& V_4=200V\end{array}$