Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by
$\frac{Q^{2}}{2 C}$
$\frac{1}{2} C V^{2}$
$\frac{1}{2 \varepsilon_{0}} \frac{V^{2}}{d^{2}} $
$\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$
Two identical capacitors are connected in parallel across a potenial difference $V$. After they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be
Work done by an external agent in separating the parallel plate capacitor is
A $2 \ \mu F$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is
If the charge on a capacitor is increased by $2C$, the energy stored in it increases by $44 \%$. The original charge on the capacitor is (in $C$ )
A piece of cloud having area $25 \times {10^6}\,{m^2}$ and electric potential of ${10^5}$ $volts$. If the height of cloud is $0.75\,km$, then energy of electric field between earth and cloud will be.....$J$