In figure , the plates of a parallel plate capacitors have unequal charges. Its capacitance is C. P is a point outside the capacitor and close to the plate of charge -Q. The distance between the plates is d. Then

Options

A. A point charge at point 'P' will experience electric force due to capacitor
B. The potential difference between the plates will be `(3Q)/(2C)`
C. The energy stored in the electric field in the region between the plates is `(9Q^(2))/(8C)`
D. The force on one plate due to the other plate is `(Q^(2))/(2pi in_(0) d^(2))`

Answer

A, B, C

Explanation

`E_(0)=(2Q)/(2 epsilon_(0)A)-(Q)/(2 epsilon_(0)A)=(Q)/(2 epsilon_(0)A)`
`E_("in") =(2Q)/(2A epsilon_())+(Q)/(2A epsilon_(0)) rArr E_("in")=(3Q)/(2A epsilon_(0))`
`E_("in")=(3)/(2)(Q)/(Cd) rArr E_("in")f=(3Q)/(2C)=V`
(ii) `F = EQ`
`F = ((2Q)/(2A epsilon_(0)))xx(-Q)=-(Q^(2))/(A epsilon_(0))`
`F = (Q^(2))/(A epsilon_(0))`
(iii) Energy `= (1)/(2) epsilon_(0) E^(2) Ad`
`= (1)/(2) epsilon_(0) ((3Q)/(2Cd))^(2) A d = (9)/(8) (Q^(2))/(C)`.