The electrostatic force between the metal plate of an isolated parallel plate capacitro `C` having charge `Q` and area `A`, is

Options

A. proportional to the square root of the distance betweenthe plates
B. linearly proportional to the distance between the plates
C. independent of the distance between the paltes
D. inversely proportional to the distance between the plates.

Answer

C

Explanation

c) As we know that, the total work done in transferring a charge to a parallel plate capacitor is given as
`W=Q^(2)/(2C)`
Where, C is the capacitance of the capacitor. We can also write a relation for work's done as,
`W= F.D`…………(ii)
Where, F is the electrostatic force between the plates of capacitor and d is the distance between the plates.
Fro m Eqs, (i) and (ii), we get
`W=Q^(2)/(2C) = Fd rArr F=Q^(2)/(2Cd)`............(ii)
As the capacitance of a parallel plate is given as
`C=(epsilon_(0)A)/d`
Substituting the value of C in Eq. (iii), we get
`F= (Q^(2)d)/(2epsilonAd) = (Q^(2))/(2epsilon_(0)A)`
this means, electrostatic force is independent of the distance between the plates.