$A$ capacitor of capacitance $C$ has charge $Q$ and energy stored in it is $W$. If the charge is increased to $3Q$, the energy stored in the capacitor $W'$ will be

$A$ parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

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:

$A$ parallel plate capacitor has a plate separation $d$ and plate area $A$. If it is charged to a potential $V$ and then disconnected from the battery,calculate the work done in increasing the separation between the plates to $2d$.

If $E$ is the electric field intensity of an electrostatic field,then the electrostatic energy density is proportional to

The charge $q$ on a capacitor varies with voltage $V$ as shown in the figure. The area of the triangle $AOB$ represents: