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

Consider a Gaussian surface as shown in figure . Two faces of this closed surface lie completely inside the condector where the electric field is zero. The flux through these faced is, therefore, zero. The other parts of the closed surface which are outside the conductor arl parallel to the electric fieeld and hence the electric field through the closed surface is, therefore, zero. From Fauss's law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B.
The distribution should be like the one shown in figure (30-W6b). To find the value of q, consider the field at a point P inside the plate A. Suppose, the surface area of the plate (one side) is A. Using the equation `E= sigma/(2epsilon_0)`, the electric field at P
due to the charge `Q_1- q= Q_1- q/ 2 A epsilon_0` (downward),
due to the charge `+q= q/ 2A epsilon_0` ( upward),
due to the charge `-q= q/2Aepsilon_0` (downward),
and due to the charge `Q_2 +a= Q_2 + a/ 2A epsilon` (upward).
The net electric field at P due to all the four charged surfaces is ( in the downward direction)
`Q_1 - a/ 2A epsilon_0 - q/2A epsilon_0 + q/2A epsilon_0- Q_2+q/2A epsilon_0.
As the point P is inside the conductor, this field should be zero. Hence,
`Q_1-q-Q_2-q=0`
or, `q= Q_-Q_2/2`. `........(1)`
Thus, `Q_1-q= Q_1=Q_2 /2` ..........(2)
and `Q_2+q = Q_1 + Q_2/2`.
Using these equation, the distribution shown in the figure (30W6) can be redrawn as in figure (30-W7).
This result is a special case of the following result. When charge conducting plates are placed parallel to each other, the twon outermost surfaces get equal charges and the facing surface get equal nad opposite charges.