What is the force of attraction between the two plates of a parallel plate capacitor? Assume that, area of each plate of the capacitor is A and one plate is charged with +Q and the other with -Q.

Explanation

Intensity of electric field at a point near a charged plate having surface density of charge `sigma` is,
`E = (sigma)/(2in_0)`
Now, `sigma = (Q)/(A)`
`:. E = (Q)/(2A in_0)`
Magnitude of the charge on the other plate of the capacitor = Q.
So, the force experienced by the plate is,
`F = QE = Q*(Q)/(2A in_0)= (Q^2)/(aA in_0)`.