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......$\mu F$.

$A$ parallel plate capacitor has circular plates of $0.08\,m$ radius and $1.0 \times 10^{-3}\,m$ separation. If a potential difference of $100\,V$ is applied,the charge on the capacitor will be:

One plate of a parallel plate capacitor is smaller than the other. What can be said about the charge on the smaller plate?

In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' is:

$A$ 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 $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

If the charge on each plate of a parallel plate capacitor is $Q$ and the magnitude of the electric field between the plates is $E$,then the force on each plate of the parallel plate capacitor will be: