Three parallel metallic plates each of area A are kept as shown in figure and charges $q_1,q_2$ and $q_3$ are given to them. Find the resulting charge distribution on the six surfaces, neglecting edge effects as usual.

Solution The plate separations do not affect the distribution of charge in this problem.
In the figure,
and
$q_b=q_1-q_a,q_d=q_2-q_c$
$q_f=q_3-q_e$
Electric field at point $P$ is zero because this point is lying inside a conductor.

At $P$, charge $q_a$ will give an electric field towards right. All other charges $q_b,q_c\dots$, etc., will give the electric field towards left. So,
$\frac{1}{2A{\epsilon }_0}\left[q_a-\left(q_1-q_a\right)-q_c-\left(q_2-q_c\right)-q_e-\left(q_3-q_e\right)\right]=0$
or
$\begin{array}{rl}2q_a-q_1-q_2-q_3& =0\\ q_a& =\frac{q_1+q_2+q_3}{2}\end{array}$
or
Similarly the condition, $E_R=0$
will give the result,
$q_f=\frac{q_1+q_2+q_3}{2}$
From here we may conclude that, half of the sum of all charges appears on each of the two outermost surfaces of the system of plates.
Further we have a condition,
$E_Q=0$
$\frac{1}{2A{\epsilon }_0}\left[q_a+\left(q_1-q_a\right)+q_c-\left(q_2-q_c\right)-q_e-\left(q_3-q_e\right)\right]=0$
or
$q_1+2q_c-q_2-q_3=0$
$\therefore q_c=\frac{q_2+q_3-q_1}{2}$
$q_b=q_1-q_a=\frac{q_1-q_2-q_3}{2}=-q_c$
Similarly, we can show that
$q_d=-q_e$
From here we can find another important result that the pairs of opposite surfaces like $\mathbf{b},\mathbf{c}$ and $d,e$ carry equal and opposite charges.