$A$ 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 $\dots \; nJ$.

$A$ $10\,\mu F$ capacitor and a $20\,\mu F$ capacitor are connected in series across a $200\,V$ supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together,and no external voltage is applied. What is the potential difference across each capacitor?

$A$ capacitor of capacitance $C_1 = 10 \mu F$ is charged using a $9 \text{ V}$ battery. It is then removed from the battery and connected to another uncharged capacitor $C_2 = 20 \mu F$ as shown in the figure. The charge on $C_2$ after equilibrium is reached is:

$A$ $2 \mu F$ 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 $\mu J$)

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:

$A$ parallel plate capacitor of capacitance $500 \ pF$ is charged with a $100 \ V$ supply. It is then disconnected from the supply and connected to another uncharged $500 \ pF$ capacitor. The electrostatic energy lost in this process is (in $\mu J$)