$A$ 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:

Three plates $A, B, C$ each of area $50 \, cm^2$ have a separation of $3 \, mm$ between $A$ and $B$ and $3 \, mm$ between $B$ and $C$. The energy stored when the plates are fully charged is (as shown in the figure):

If the distance between the plates of a capacitor having capacity $C$ and charge $Q$ is doubled,then the work done will be:

If the potential of a capacitor having capacity of $6\,\mu F$ is increased from $10\,V$ to $20\,V$,then the increase in its energy will be:

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.

If the charge on a body is increased by $2 \ C$,the energy stored in it increases by $21 \%$. The original charge on the body in coulomb is: