Question

$A$,$B$ and $C$ are three identical, infinitely long , parallel conducting plates placed as shown in the figure. If $q_1$ and $q_2$ be the charges present on the surfaces left of $A$ and right of $B$ then find the value of $q_1$ and $q_2$

• A. $q_1=\frac{3Q}{2}$, $q_2=\frac{-Q}{2}$
• B. $q_1=\frac{-3Q}{2}$, $q_2=\frac{Q}{2}$
• C. $q_1=\frac{3Q}{2}$, $q_2=\frac{Q}{2}$
• D. $q_1=\frac{3Q}{2}$, $q_2=\frac{3Q}{2}$

Answer 1

The correct option is C $q_1=\frac{3Q}{2}$, $q_2=\frac{Q}{2}$

Let conducting plate $A$ has charge$"-Q"$ , conducting plate $B$ has charge $"3Q"$ and conducting plate $C$ has charge $"Q"$.

Let us assume the charge distribution on all the plates is as shown in the figure and $P$ be a point inside the the conducting plate $A$.


To find charges $q_1$ and $q_2$

Net field at point $\text{P}$ due to charges on each surface:

$E_P=\frac{(-Q-x)}{2A{\epsilon }_0}-(\frac{x}{2A{\epsilon }_0}+\frac{3Q}{2A{\epsilon }_0}+\frac{Q}{2A{\epsilon }_0})$

Since, $\text{P}$ is inside the conductor.
$E_P=0$

$\Rightarrow \frac{(-Q-x)}{2A{\epsilon }_0}-(\frac{x}{2A{\epsilon }_0}+\frac{3Q}{2A{\epsilon }_0}+\frac{Q}{2A{\epsilon }_0})=0$

$\Rightarrow x=\frac{-5Q}{2}$

So, $(-Q-x)=(-Q+\frac{5Q}{2})$

$\therefore q_1=(-Q-x)=\frac{3Q}{2}$

Since, here electric field inside the Gaussian surface is zero , the net charge enclosed by the Gaussian surface must be zero. So,

$y=-x$

$\therefore y=-(-\frac{5Q}{2})=\frac{5Q}{2}$

Now, from conservation of charge, $q_2+y=3Q$

$\Rightarrow q_2=3Q-y$

Substituting the value of $y$ we get,

$q_2=3Q-\frac{5Q}{2}$

$\Rightarrow q_2=\frac{Q}{2}$

Hence option (c) is the correct answer.

Why this question? $\text{Key Point:}$ The net electric field at any point is independent of charge distribution on the plate, it only depends on the total charge on a plate.