Two hollow metallic spheres of radii $20 \ cm$ and $10 \ cm$ each carry a charge of $150 \ \mu C$. If they are connected by a conducting wire,the common potential is ...... $V$.

Two parallel plate capacitors of capacitances $20 \mu F$ and $30 \mu F$ are charged to potentials of $30 \, V$ and $20 \, V$ respectively. If the similarly charged plates are connected together,then the common potential difference will be: (in $, V$)

Two metallic spheres of radii $1 \, cm$ and $3 \, cm$ are given,with charges of $-1 \times 10^{-2} \, C$ and $5 \times 10^{-2} \, C$ respectively. If these are connected by a conducting wire,the final charge on the bigger sphere is

$A$ parallel plate capacitor of capacitance $10 \mu F$ is charged by a $220 \text{ V}$ supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance $12 \mu F$. The loss of electrostatic energy in this process is (in $\text{ mJ}$)

Two spheres $A$ and $B$ of radii $4 \text{ cm}$ and $6 \text{ cm}$ are given charges of $80 \mu\text{C}$ and $40 \mu\text{C}$ respectively. If they are connected by a fine wire,the amount of charge flowing from one to the other is

In the circuit shown,$C_1 = 6\,\mu F, C_2 = 3 \,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge (in $\mu C$) finally on $C_2$?