$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another identical uncharged capacitor is connected in parallel with it. The total energy of the system of capacitors is . . . . . . .

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 $pF$)

$A$ capacitor of capacitance $900\,\mu F$ is charged by a $100\,V$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of the uncharged capacitor is connected to the positive plate and the other plate of the uncharged capacitor is connected to the negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-2}\,J$. The value of $x$ is $..............$

Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

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:

In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]