Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

Energy $E$ is stored in a parallel plate capacitor $C_1$. An identical uncharged capacitor $C_2$ is connected to it,kept in contact with it for a while,and then disconnected. The energy stored in $C_2$ is

The circuit shows two capacitors $A$ and $B$ of capacitances $C$ and $2C$ respectively. When they are fully charged,the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then:
$(a)$ Charge on $A$ is $\frac{4CE}{9}$
$(b)$ Charge on $B$ is $\frac{8CE}{9}$
$(c)$ Loss of energy in this process is $\frac{CE^2}{3}$
The correct statement$(s)$ is/are:

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

$A$ $2\,\mu F$ capacitor is charged to a potential of $10\,V$. Another $4\,\mu F$ capacitor is charged to a potential of $20\,V$. 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 $\mu J$?