$A$ $40 \, \mu F$ capacitor in a defibrillator is charged to $3000 \, V$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2 \, ms$. The power delivered to the patient is ...... $kW$.

$A$ 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:

$A$ capacitor of capacitance $10 \mu F$ is charged to $10 \text{ V}$. The energy stored in it is (in $\mu J$)

The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection,if the plates of the capacitor are separated so that the distance between them is twice the original distance,then the electrostatic energy becomes

Work done in placing a charge of $8 \times 10^{-18} \text{ C}$ on a capacitor of capacitance $800 \mu\text{F}$ is . . . . . . .

Capacity of a capacitor is $48 \mu F$. When it is charged from $0.1 C$ to $0.5 C$,the change in the energy stored is: