Force of attraction between the plates of a parallel plate capacitor is

Options

A. `(q^(2))/(epsi_(0)AK)`
B. `(q^(2))/(2epsi_(0)AK)`
C. `(q^(2))/(2epsi_(0)A)`
D. `(q^(2))/(2epsi_(0)A^(2)K)`

Answer

B

Explanation

Consider a parallel plate capacitor, of area A, plate separation d and filled with a medium of dielectric constant K.
The electric field near a non-conducting plane sheet (P) having uniform surface charge density `sigma` is given by
`E=(sigma)/(2epsi_(0)K)`
`therefore` The force on the plate Q due to the charged plate P is
`F=qE=q*((sigma)/(2epsi_(0)K))=(q)/(2epsi_(0)K)((q)/(A))" "therefore sigma=(q)/(A)`
`therefore F=(q^(2))/(2epsi_(0)AK)`
Second Method:
The energy of a parallel plate capacitor
`U=(1)/(2) (q^(2))/(C)=(q^(2)d)/(2*KA epsi_(0))" "becauseC=(KA epsi_(0))/(d)`
`therefore`If the plate separation is increased by `Deltax`, then increases in energy `DeltaU=(1)/(2)(q^(2)Deltax)/(KA epsi_(0))` . . . (1)
[Except `Deltax` all other quantities remain constant]
If F is the force between the plates, thenn the work done in increasing the plate separation by `Deltax` is
`dW=Fdx` . . (2)
but `DeltaU=dW" "therefore`From (1) and (2)
`therefore Fdx=(1)/(2)(q^(2)Deltax)/(KA epsi_(0))" "therefore F=(1)/(2)(q^(2))/(KA epsi_(0))`