Question

Three identical metal plates with large surface areas are kept parallel to each other as shown in the figure (30-E8). The leftmost plate is given a charge Q, the rightmost a charge −2Q and the middle one is kept neutral. Find the charge appearing on the outer surface of the rightmost plate.

Answer 1

Consider the Gaussian surface as shown in the figure.
Let the charge on the outer surface of the left-most plate be q. Thus, the charges on the plates are distributed as shown in the diagram.

The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.

Let the surface area of the plates be A.

Electric field at point P due to the charges on plate X:
Due to charge (+Q $-$ q) is $\frac{Q-q}{2A{\in }_0}$ in the right direction
Due to charge (+q) is $\frac{q}{2A{\in }_0}$ in the right direction

Electric field at point P due to charges on plate Y:

Due to charge ($-$q) is $\frac{q}{2A{\in }_0}$ in the right direction
Due to charge (+q) is $\frac{q}{2A{\in }_0}$ in the left direction


Electric field at point P due to charges on plate Z:
Due to charge ($-$q) is $\frac{q}{2A{\in }_0}$ in the right direction
Due to charge ($-$2Q + q) is $\frac{2Q-q}{2A{\in }_0}$ in the right direction

The net electric field at point P:
$\frac{Q-q}{2A{\in }_0}$ + $\frac{q}{2A{\in }_0}$ $-$ $\frac{q}{2A{\in }_0}$ $-$ $\frac{q}{2A{\in }_0}$ + $\frac{q}{2A{\in }_0}$ + $\frac{2Q-q}{2A{\in }_0}$ = 0

$\frac{Q-q}{2A{\in }_0}$ + $\frac{2Q-q}{2A{\in }_0}$ = 0
$$\begin{gathered} Q-q+2Q-q=0 \\ 3Q-q=0 \\ q=\frac{3Q}{2} \end{gathered}$$

Thus, the charge on the outer plate of the right-most plate = $-2Q+q=-2Q+\frac{3Q}{2}=-\frac{Q}{2}$