Charges `2q` and `-3q` are given to two identical metal plates of area of cross section `A`. the distance between the plates is `d`. Find the capacitance and potential difference between the plates.

Answer

A, B, D

Explanation

Charge on outercome surfaces
`q_(t otal)/2=(2q-3q)/2=-q/2`
Hence, charge on different faces are as shown below

electric field and hence potentiasl difference between the two plates is due to `+-2.5q`.
`PD=Ed`
`=(sigma/epsilon_0)d=((2.5qd)/(epsilon_0A))`
Capacitance `C=(epsilon_0A)/d`