$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

$A$ $0.2 \, F$ capacitor is charged to $600 \, V$ by a battery. After removing the battery,it is connected to another uncharged parallel plate capacitor of $1 \, F$. The potential decreases to $V$ equal to:

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

Two insulated spheres of radii $R_1$ and $R_2$ having charges $Q_1$ and $Q_2$ respectively are connected to each other. There is:

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