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

Six metallic plates,each with a surface area $A$ on one side,are placed at a distance $d$ from each other. The alternate plates are connected to points $P$ and $Q$ as shown in the figure. The capacitance of the system is:

Two parallel plate capacitors with different plate separations but the same capacitance are connected in series to a battery. Both capacitors are filled with air. The quantity that is $NOT$ the same for both the capacitors when they are fully charged 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 $\epsilon_0 = 9 \times 10^{-12} \ F/m, \pi = 22/7$) . . . . . . $\mu m$.

Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$,the potential difference $V$ between the plates is

What is the area of the plates of a capacitor with a capacitance of $3 \ F$? The distance between the two plates is $5 \ mm$.