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

Options

A. `(C^(2)V^(2))/(2d^(2))`
B. `(C^(2)V^(2))/(2d)`
C. `(CV^(2))/(2d)`
D. `(CV^(2))/(d)`

Answer

B, C, D

Explanation

Electric field produced by one plate `E_(1) = (sigma)/(2 epsilon_(0))`
Force on second plte with charge `sigmaA`
`F = E_(1) ( sigmaA)`
`:. F = (sigma^(2)A)/(2 epsilon_(0))` [ From equation (1) ]
But `sigma= (Q)/(A)`
`:. F=((Q^(2))/A^(2).A)/(2epsilon_(0))=(Q^(2))/(2epsilon_(0)A)=(Q^(2))/(2Cd)[ because epsilon_(0)A = Cd]`
`:. F =(C^(2)V^(2))/(2Cd) " " [ because Q = CV]`
` :. F = (CV^(2))/(2d)`