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.

$A$ conducting sphere of radius $R$ carrying a charge $q$ is joined by a conducting wire to a conducting sphere of radius $2R$ carrying a charge $-2q$. The charge flowing between them will be:

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

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:

Two identical capacitors are connected in parallel across a potential difference $V$. After they are fully charged,the positive plate of the first capacitor is connected to the negative plate of the second,and the negative plate of the first is connected to the positive plate of the other. The loss of energy will be

Two capacitors of $100 \mu F$ and $50 \mu F$ are connected in parallel. If the potential difference across $100 \mu F$ is $20 \text{ V}$ and across $50 \mu F$ is $40 \text{ V}$,then the common potential of the parallel combination will be (assuming same polarities of the capacitors are connected together).