$A$ parallel plate capacitor is charged by a battery until the potential difference between the plates equals the electromotive force of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

Given $C_1 = 2\ \mu F$,$C_2 = 6\ \mu F$,and $C_3 = 4\ \mu F$. What is the total energy of the system?

$A$ capacitor is charged by a battery and the energy stored in it is $U$. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is:

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

For a capacitor with area $A$ and potential difference $V$ separated by a distance $d$,the energy per unit volume is:

The plates of a parallel plate capacitor have an area of $90 \,cm^{2}$ each and are separated by $2.5 \,mm$. The capacitor is charged by connecting it to a $400 \,V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates,and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.