$A$ capacitor of capacitance $10\,\mu F$ has been charged to $100\,V$. It is now connected to another uncharged capacitor in parallel. The common potential becomes $40\,V$. The capacitance of the other capacitor is......$\mu F$.

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

$A$ condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. The energy dissipated in the process is

Two capacitors of capacitance $2 \ \mu F$ and $5 \ \mu F$ are charged to $2 \ V$ and $10 \ V$ respectively. Find the ratio of their charges after they are connected by a wire.

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 $, V$)

$A$ $2 \mu F$ capacitor is connected to a $50 \ V$ supply,and a $3 \mu F$ capacitor is connected to a $100 \ V$ supply. After removing the batteries,if the two plates of the same type of charge are connected to each other,the new potential difference is . . . . . . $V$.