The energy required to charge a capacitor of $5\,\mu F$ by connecting a $d.c.$ source of $20\,kV$ is......$kJ$.

When a parallel plate capacitor is fully charged by a battery,the potential difference between the two plates becomes equal to the electromotive force $(EMF)$ of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

$A$ parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A,$ the energy stored in the capacitor is

$A$ parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates,the time rate of change of electrostatic energy of the capacitor is proportional to

$A$ 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).

$A$ capacitor of capacity $50\,\mu F$ is charged to $10\;V$. Its energy is equal to: