Question

Two conducting plates A and B are placed parallel to each other at a small distance between them. Plate A is given a charge $q_1$ and plate B is given a charge $q_2$. Then :

• A. the outer surfaces of A and B (not facing each other) get no charge
• B. the inner surfaces of A and B (facing each other) get all the charge
• C. the inner surfaces of A and B (facing each other) get equal and opposite charge of magnitude $|\frac{q_1-q_2}{2}|$
• D. the outer surfaces of A and B (not facing each other) get charge of the same polarity and of magnitude $|\frac{q_1+q_2}{2}|$

Answer 1

Answer: (C), (D)

The correct options are
C the inner surfaces of A and B (facing each other) get equal and opposite charge of magnitude $|\frac{q_1-q_2}{2}|$
D the outer surfaces of A and B (not facing each other) get charge of the same polarity and of magnitude $|\frac{q_1+q_2}{2}|$

Consider the charges to be distributed as shown in the figure.

As the total amount of charge on plate $1$ is $q_1$, we have $Q_1+Q_2=q_1$ -----(1)

Similarly, on the second plate, $Q_3+Q_4=q_2$ --------(2)

Consider the points $A$ and $B$. Both the points are within the conducting plates. Thus, Electric field at $A$ and $B$ is Zero.

Electric field due to a sheet of charge is $E=\frac{Q}{2A{ϵ}_0}$ and is normal to the plate and away from it.

Electric field at $A$ is given by $E_A=\frac{1}{2A{ϵ}_0}(Q_1\left(\hat{i}\right)+Q_2(-\hat{i})+Q_3(-\hat{i})+Q_4(-\hat{i}))=\frac{Q_1-Q_2-Q_3-Q_4}{2A{ϵ}_0}\hat{i}$

Thus, $Q_1-Q_2-Q_3-Q_4=0$ ------(3)

Similarly, electric field at $B$ is $E_B=\frac{Q_1+Q_2+Q_3-Q_4}{2A{ϵ}_0}\hat{i}$

Thus, $Q_1+Q_2+Q_3-Q_4=0$ -------(4)

Adding (3) and (4), we have $Q_1-Q_4=0\Rightarrow Q_1=Q_4$

Subtracting (3) and (4), we have $Q_2+Q_3=0\Rightarrow Q_2=-Q_3$

Substituting in (1) and (2), we have,

$Q_1=\frac{q_1+q_2}{2},Q_2=\frac{q_1-q_2}{2},Q_3=\frac{q_2-q_1}{2},Q_4=\frac{q_1+q_2}{2}$