A $700\,pF$ capacitor is charged by a $50\,V$ battery. The electrostatic energy stored by it is
$17.0 \times {10^{ - 8}}\,J$
$13.6 \times {10^{ - 9}}\,J$
$9.5 \times {10^{ - 9}}\,J$
$8.7 \times {10^{ - 7}}\,J$
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 area of each plate is $A,$ the energy stored in the capacitor is
A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be
A capacitor $C$ is charged to a potential difference $V$ and battery is disconnected. Now if the capacitor plates are brought close slowly by some distance :
A condenser of capacity $50\,\mu F$ is charged to $10\;volts$. Its energy is equal to
Change $Q$ on a capacitor varies with voltage $V$ as shown in the figure, where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents