$A$ $20 \text{ F}$ capacitor is charged to $5 \text{ V}$ and isolated. It is then connected in parallel with an uncharged $30 \text{ F}$ capacitor. The decrease in the energy of the system will be: (in $\text{ J}$)

$A$ condenser having a capacity of $6\,\mu F$ is charged to $100\, V$ and is then joined to an uncharged condenser of $14\,\mu F$ and then removed. The ratio of the charges on $6\,\mu F$ and $14\,\mu F$ and the potential of $6\,\mu F$ will be

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:

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

$A$ conducting sphere $S_1$ of radius $r_1 = 3 \text{ cm}$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2 = 2 \text{ cm}$. Before they are connected,$S_1$ carries a charge of $10 \text{ units}$. The electric potential at a point which is at a distance $4 \text{ cm}$ from the centre of $S_1$ and a distance $3 \text{ cm}$ from the centre of $S_2$ is:

$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .