Two parallel air capacitors have their plate areas $100 \,cm^2$ and $500 \,cm^2$, respectively. They have the same charge and potential. If the distance between the plates of the first capacitor is $0.5 \,mm$, the distance between the plates of the second capacitor is (in $\,cm$)

$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

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

The charge on a parallel plate capacitor of capacity $C$ is $Q$. The electric field intensity between its two plates,which are separated by a distance $t$,is:

What is the area of the plates of a $3\,F$ parallel plate capacitor,if the separation between the plates is $5\,mm$?

Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).