$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

Two insulated metallic spheres of $3\,\mu F$ and $5\,\mu F$ capacitances are charged to $300\, V$ and $500\, V$ respectively. The energy loss,when they are connected by a wire,is (in $,J$)

When two isolated conductors $A$ and $B$ are connected by a conducting wire,positive charge will flow from:

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

$A$ $2\,\mu F$ capacitor has a potential difference of $200\;V$ across its terminals. It is disconnected from the battery and then another uncharged capacitor is connected in parallel to it. If the potential difference becomes $20\;V$,the capacitance of the second capacitor is.......$\mu F$.

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using a battery of $emf$ $V$ volt. Now,disconnecting $a$ and $b$,the terminals $b$ and $c$ are connected. Due to this,what will be the percentage loss of energy?