The energy of a charged capacitor is given by the expression ($q$ = charge on the conductor and $C$ = its capacity).

If a $12 \ pF$ capacitor is connected to a $50 \ V$ battery,then the electrostatic energy stored in the capacitor is . . . . . . .

$A$ 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 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 plates is:

$A$ parallel plate capacitor has plate area $A$ and separation between plates is $d$. It is charged to a potential difference of $V_0$ volt. The charging battery is then disconnected and plates are pulled apart to three times the initial distance. The work done to increase the distance between the plates is $(\varepsilon_0 = \text{permittivity of free space})$

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