Two conducting plates A and B are placed parallel to each other. A is given a charge. `Q_1` and B a charge `Q_2`. Find the distribution of charges on the four surface.

Explanation

(i) When they are at large distance from each other.

Electric field inside the conductor is zero. This is the case when both the surface have same surface charge density. Thus, the charge distributes equality on both sides.
(ii) When they are brought closer to each other, the electric field is the superposition field. But net field inside the conductor roust be zero. Let the cahrge distribution in the figure. Charge densities at different surface will be
`sigma_(1)= (Q_(1)-q_(1))/(A)`
`sigma_(2)= (q_(1))/(A)`
`sigma_(3)= (q_(2))/(A)`
`sigma_(4)= (Q_(2) - q_(2))/(A)`
Region I is inside a conductor , therefore `E =0`
`implies (sigma_(1)-sigma_(2)-sigma_(3)-sigma_(4))/(epsilon_(0))=0`
`implies sigma_(1)=sigma_(2)+sigma_(3)+sigma_(4)` ....(i)

Similarly, for region III, E=0,
`= sigma _(4)-sigma_(1)+sigma_(2)-sigma_(3)=0`
or `sigma_(4) = sigma_(1) + sigma_(2)+sigma_(3)` (ii)
From (i) and (ii)
`sigma_(2)+ sigma_(3)=0 implies q_(1)= -q_(2)`
Also, `sigma_(1) = sigma_(4) implies Q_(1)-q_(1) = Q_(2) -q_(2)`
Put `q_(2) = -q_(1) implies Q_(1) - q_(1) = Q_(2) + q_(1)`
or, `q_(1) = (Q_(1)-Q_(2))/(2) implies q_(2) = - ((Q_(1)-Q_(2))/(2))`
Thus, the charge distribution is as shown

where `q= ( Q_(1) - Q_(2))/(2)`