Question

An isolated large conducting plate having area $\text{A}$ and total charge $Q$ is placed in a uniform electric field $\overrightarrow{E}$ as shown in the figure. If $q_1$ and $q_2$ are the charges appearing on the two sides of the plate, then

• A. $q_1=\frac{Q}{2}+EA{\epsilon }_0$ , $q_2=\frac{Q}{2}-EA{\epsilon }_0$
• B. $q_1=\frac{Q}{3}$, $q_2=\frac{Q}{2}-EA{\epsilon }_0$
• C. None of the above
• D. $q_1=\frac{Q}{2}-EA{\epsilon }_0$, $q_2=\frac{Q}{2}+EA{\epsilon }_0$

Answer 1

The correct option is D $q_1=\frac{Q}{2}-EA{\epsilon }_0$, $q_2=\frac{Q}{2}+EA{\epsilon }_0$

Given,
Net charge on the plate$=Q$
Area of the plate$=A$

Let us consider a point $\text{P}$ inside the plate as shown in figure.


As we know that, electric field at point $\text{P}$ inside the conductor is zero. So,
$E_{net}=0$

Electric field at point $\text{P}$ due to charge $q_1$ is $\frac{q_1}{2A{\epsilon }_o}$ as shown in figure.

Similary,
Electric field at point $\text{P}$ due to charge $q_2$ is $\frac{q_2}{2A{\epsilon }_o}$ as shown in figure.

Thus, net electric field at point $\text{P}$,

$E_{net}=\frac{q_1}{2A{\epsilon }_o}+E-\frac{q_2}{2A{ϵ}_0}$

Since, $E_{net}=0$, so we have

$\frac{q_1}{2A{\epsilon }_o}+E-\frac{q_2}{2A{ϵ}_0}=0.......\left(1\right)$

Also, from conservation of charge,
$q_1+q_2=Q.......\left(2\right)$

From (1) and (2), we have

$\frac{q_1}{2A{\epsilon }_o}+E-\frac{Q-q_1}{2A{ϵ}_0}=0$

$\Rightarrow \frac{q_1}{2A{\epsilon }_o}+E-\frac{Q}{2A{ϵ}_0}+\frac{q_1}{2A{ϵ}_0}=0$

$\Rightarrow \frac{q_1}{A{\epsilon }_o}=\frac{Q}{2A{\epsilon }_o}-E$

$\therefore q_1=\frac{Q}{2}-EA{\epsilon }_o$

From equation (2),
$q_2=\frac{Q}{2}+EA{\epsilon }_0$

Hence option (a) is the correct answer.