$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:

Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

Two metal spheres of capacitance $C_1$ and $C_2$ carry some charges. They are put in contact and then separated. The final charges $Q_1$ and $Q_2$ on them will satisfy

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

$A$ 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 $\text{ mJ}$)

$A$ $2 \mu F$ capacitor is charged to a potential $V_1 = 10 \ V$. Another $4 \mu F$ capacitor is charged to a potential $V_2 = 20 \ V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit? (in $\mu J$)