Question

Three metallic plates each of area A are kept as shown in figure and charges $Q_1, Q_2, Q_3$ are given to them to find the resulting charge distribution on the six surfaces, neglecting edge effect?

Answer 1

In the system of large isolated parallel metal plates, the charges are distributed in such a way that always the facing surfaces of the plates carry equal and opposite charges and outer surfaces of the system carry equal charges.

Complete step by step answer:
Given: Area of each plate is A.
Charges on the three metallic plates are Q1, Q2 and Q3 respectively.
Let the charges distributed on each plate be $q_1, q_2, q_3, q_4, q_5$ and $q_6$ as shown in the figure below

According to redistribution and charge conservation, we have
${q_1} = {q_6} = \dfrac{{{Q_1} + {Q_2} + {Q_3}}}{2}$, ${q_3} = - {\text{ }}{q_2}$and ${q_4} = - {\text{ }}{q_5}$
${q_2}{\text{ }} = {\text{ }}{Q_1} - {\text{ }}{q_1}$
$\Rightarrow {q_2}{\text{ }} = {\text{ }}{Q_1} - {\text{ }}{q_1} = \;{Q_1} - \dfrac{{{Q_1} + {Q_2} + {Q_3}}}{2}$
$\therefore {q_2}{\text{ }} = \dfrac{{{Q_1} - {Q_2} - {Q_3}}}{2}$.

Note: Whenever a charge is given to a conducting body, or on a metallic body, the charge will spread on the outer surface of the body due to the mutual repulsion. The charge is automatically distributed in such a way that the net electric field intensity at every interior point of the body after the distribution becomes zero. The final charge distribution on all the plates can be directly calculated by using two facts. One is that all facing surfaces of the plates have equal and opposite polarity charges and the second one is that the total charges on the outermost surface of side plates of the system are always equal and half of the sum of all the charges on the plates. This second statement can also be deduced by considering zero electric fields at the interior points of the side metal plates.