If potential difference across a capacitor is changed from $15 \ V$ to $30 \ V,$ work done is $W.$ The work done,when potential difference is changed from $30 \ V$ to $60 \ V,$ will be $.......W$

$A$ 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:

$A$ capacitor is charged to $200\,V$. It has $0.1\,C$ of charge. When it is discharged,the energy released will be $J$.

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the energy stored in the capacitor.

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

Three plates $A, B, C$ each have an area of $50 \, cm^2$ and the distance between them is $3 \, mm$. Find the energy stored in the system when the plates are fully charged.