$A$ $2 \mu F$ capacitor is connected to a $50 \ V$ supply,and a $3 \mu F$ capacitor is connected to a $100 \ V$ supply. After removing the batteries,if the two plates of the same type of charge are connected to each other,the new potential difference is . . . . . . $V$.

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 $C_2$?

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12\, \mu C$ and $-3\, \mu C$ charges,respectively,and are at a large distance from each other. They are now connected by a conducting wire. $A$ long time after this is done,the charges on $S_{1}$ and $S_{2}$ are respectively:

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging batteries are then removed,and the capacitors are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

$A$ condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. The energy dissipated in the process is

Two insulated metallic spheres of $3\,\mu F$ and $5\,\mu F$ capacitances are charged to $300\, V$ and $500\, V$ respectively. The energy loss,when they are connected by a wire,is (in $,J$)