Consider the circuit shown in the figure below. All the resistors are identical. The charge stored in the capacitor,once it is fully charged is

Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2},$ respectively. The electric fields $E_{1}$ (on $S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that $E_{1} / E_{2} = R_{1} / R_{2}.$ Then the ratio $V_{1} / V_{2}$ of the electrostatic potentials on each sphere is:

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

Five capacitors of capacitances $C_1 = C_2 = C_3 = C_4 = 10 \mu F$ and $C_5 = 2.5 \mu F$ are connected as shown,along with a battery of $50 \ V$. Find the equivalent capacitance and the charge on the capacitors.

In the circuit shown in the figure,$C_1 = C_2 = 2 \mu F$. Find the charge stored in the capacitors.

Three charged capacitors,$C_1 = 17 \ \mu F$,$C_2 = 34 \ \mu F$,$C_3 = 41 \ \mu F$ and two open switches,$S_1$ and $S_2$ are assembled into a network with initial voltages and polarities,as shown in the figure. The final status of the network is attained when the two switches,$S_1$ and $S_2$ are closed. In the figure,the final charge on capacitor $C_3$ in $mC$ is closest to: