Two capacitors $C_1$ and $C_2$ are charged to $120 \ V$ and $200 \ V$ respectively. It is found that connecting them together the potential on each one can be made zero. Then

Two parallel plate capacitors of capacitances $20 \mu F$ and $30 \mu F$ are charged to potentials of $30 \, V$ and $20 \, V$ respectively. If the similarly charged plates are connected together,then the common potential difference will be: (in $, V$)

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch and a resistor as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on each capacitor will be:

$A$ 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 $\text{ mJ}$)

$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

The energy stored in the capacitor shown in figure $(a)$ is $4.5 \times 10^{-6} \ J$. If the battery is disconnected and the capacitor is connected to another uncharged capacitor of $900 \ pF$ as shown in figure $(b)$,what will be the total energy of the system?