Question

Two parallel conducting plates have charges on different identical surfaces as shown in the figure. Which of the following statements is/are correct?

• A. $$q_1=q_3$$
• B. $$q_1=-q_3$$
• C. $$q_2=q_4$$
• D. $$q_2=\frac{q_1+q_4}{2}$$

Answer 1

Answer: (B), (C)

$$q_2=q_4$$

Let $q_1$ be the charge reciding on surface $\left(1\right)$, $q_2$ be the charge on surface $\left(2\right)$, $q_3$ be the charge on surface $\left(3\right)$ and $q_4$ be the charge on surface $\left(4\right)$.

Let us take a Gaussian surface as shown in the figure below

Since, no field lines will cut the surfaces $\text{AB}$ and $\text{CD}$ (as they are parallel to the field lines), so flux through $\text{AB}$ and $\text{CD}$ surface will be zero.

Also, electric field inside the conductor is zero, so the electric flux through $\text{BC}$ and $\text{DA}$ will also be zero.

So, from gauss theorem we can conclude that,

${\varphi }_{net}=\frac{q_{in}}{{\epsilon }_0}=0$

$\Rightarrow q_{in}=0$

From figure, charge enclosed by the Gaussian surface,
$q_{in}=q_1+q_3=0$

$\Rightarrow q_1=-q_3$

Let us choose a point $\text{P}$ on the surface $\text{AC}$:

As electric field inside the conductor is zero, so $|\overrightarrow{E}{|}_p=0$

$\therefore E_2=E_1+E_3+E_4$

where, $E_1,E_2,E_3$ and $E_4$ are electric fields due to charges present on face $1,2,3$ and $4$ respectively.

Since electric field formula due to charge on plate surface,$$E=\frac{Q}{2A{\epsilon }_0}$$ So, applying above formula for each plate, we get

$\frac{q_2}{2A{\epsilon }_0}=\frac{q_1}{2A{\epsilon }_0}+\frac{q_3}{2A{\epsilon }_0}+\frac{q_4}{2A{\epsilon }_0}$

where, $A$ is the surface area of plate.

$\Rightarrow q_2=q_1+q_3+q_4$

As, $q_1=-q_3$ we get, $q_2=-q_3+q_3+q_4$

$\Rightarrow q_2=q_4$

Hence, options (b) and (c) are the correct answers.

$\text{Note:}$ Students are adviced to memorise these two results, as these are extremly important to solve many other problems.