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`. Prove that the charges on the inner surfaces are of equal magnitude and opposite sign.

Explanation

Consider a Gaussian surface as shown in figure. Two faces of this closed surface life completely inside the conductor where the electric field is zero. The flux throught these faces is therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss's law, the total charge inside this closed surface should be zero. the charge on the inner surface of a should 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. 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//(2 epsi_(0))`, the electric field at P
Due to the charge `Q_(1)-q=(Q_(1)-q)/(2Aepsi_(0))` (downward)
Due to the charge `+q=q/(2 Aepsi_(0))` (upward),
Due to the charge `-q=q/(2 a epsi_(0))` (downward),
and due to the charge `Q_(2)+q=(Q_(2)+q)/(2Aepsi_(0))` (upward).
The net electric field at P due to all the four charged surfaces is (in the downward direction)
`E_(p)=(Q_(1)-q)/(2 A epsi_(0))-q/(2Aepsi_(0))+q/(2Aepsi_(0))-(Q_(2)+q)/(2Aepsi_(0))`
As the point P is inside the conductor, this field should be zero. Hence,
`Q_(1)-q-q+q-Q_(2)-q=0` or, `q=(Q_(1)-Q_(2))/(2)`
This result is a special case of the following result. When charged conducting plates are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal opposite charges.