If the charge on the capacitor is increased by $3 \ C$,the energy stored in it increases by $44 \%$. The original charge on the capacitor is (in $C$)

The energy stored in a capacitor of capacitance $10 \mu F$ when charged to a potential of $6 \text{ kV}$ is: (in $\text{ J}$)

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ($\epsilon_0 =$ permittivity of free space).

$A$ parallel plate capacitor has a uniform electric field $E \ (V/m)$ in the space between the plates. If the distance between the plates is $d \ (m)$ and the area of each plate is $A \ (m^2)$,then the energy (joules) stored in the capacitor is:

If the potential difference across a capacitor is increased from $5 \,V$ to $15 \,V$, then the ratio of final energy to initial energy stored in the capacitor is

$A$ capacitor having a capacity of $2.0 \; \mu F$ is charged to $200 \; V$ and then the plates of the capacitor are connected to a resistance wire. The heat produced in joules will be: