$A$ battery is used to charge a parallel plate capacitor until the potential difference between the plates becomes equal to the e.m.f. of the battery. The ratio of the energy stored in the capacitor to the work done by the battery will be

The amount of work done in increasing the voltage across the plates of a capacitor from $5 V$ to $10 V$ is $W$. The work done in increasing it from $10 V$ to $15 V$ will be

Two capacitors,one of capacity $\frac{C}{2}$ and the other of capacity $C$,are connected to a battery of voltage $V$ as shown. The work done in charging both the capacitors fully is

$A$ parallel plate air capacitor has capacity $C$,distance of separation between plates is $d$,and potential difference $V$ is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is

$A$ parallel plate capacitor has a uniform electric field $\overrightarrow{E}$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A$,the energy stored in the capacitor is: ($\varepsilon_{0} =$ permittivity of free space)

$A$ capacitor of capacitance $6 \, \mu F$ is charged up to $100 \, V$. The energy stored in the capacitor is ........ $J$.