$A$ parallel plate capacitor has an electric field of $10^5 \, V/m$ between the plates. If the charge on the capacitor plate is $1 \, \mu C$,the force on each capacitor plate is ...... $N$.

$A$ capacitor is connected to a battery. What happens to the force of attraction between the plates when the separation between them is halved?

$A$ parallel plate capacitor consisting of two circular plates of radius $10 \ cm$ is being charged by a constant current of $0.15 \ A$. If the rate of change of potential difference between the plates is $7 \times 10^8 \ V/s$,then the integer value of the distance between the parallel plates is $—$ (Take $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

$A$ parallel plate capacitor is connected to a battery,which maintains a constant potential difference. If the plates of the capacitor are moved further apart,the electric field intensity...

$A$ parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles,then

The capacity of a parallel plate capacitor is $15\,\mu F$,when the distance between its plates is $6\,cm$. If the distance between the plates is reduced to $2\,cm$,then the capacity of this parallel plate capacitor will be......$\mu F$.