Question

Two parallel metallic plates have surface charge densities ${\sigma }_1$ and ${\sigma }_2$ as shown in figure. Match the following.

Table-1 Table-2

$\left(\text{A}\right)$ If ${\sigma }_1+{\sigma }_2=0$	$\left(\text{P}\right)$ Electric field in region $\text{III}$ is towards right.
$\left(\text{B}\right)$ If ${\sigma }_1+{\sigma }_2>0$	$\left(\text{Q}\right)$ Electric field in region $\text{I}$ is zero.
$\left(\text{C}\right)$ If ${\sigma }_1+{\sigma }_2<0$	$\left(\text{R}\right)$ Electric field in region $\text{I}$ is towards right.
$\left(\text{S}\right)$ None.
$\left(\text{T}\right)$ Nothing can be said.

• A. $\text{A}\to \text{P},\text{B}\to \text{Q},\text{C}\to \text{R}$
• B. $\text{A}\to \text{Q},\text{B}\to \text{P},\text{C}\to \text{R}$
• C. $\text{A}\to \text{R},\text{B}\to \text{Q},\text{C}\to \text{P}$
• D. $\text{A}\to \text{R},\text{B}\to \text{P},\text{C}\to \text{Q}$

Answer 1

The correct option is B $\text{A}\to \text{Q},\text{B}\to \text{P},\text{C}\to \text{R}$

$\text{CASE-I}:$

For , ${\sigma }_1+{\sigma }_2=0$ we get, ${\sigma }_1=-{\sigma }_2$

That implies, Electric fields in regions $\text{I}$ and $\text{III}$ are zero.

So , $\text{A}\to \text{Q}$

$\text{CASE-II}:$

For , ${\sigma }_1+{\sigma }_2>0$ we get, ${\sigma }_1>-{\sigma }_2$

Thus, Charge on the left plate is greater compared to the charge on the right plate.

So, Electric field in the region $\text{I}$ is towards left and in the region $\text{III}$ is towards right.

$\therefore \text{B}\to \text{P}$

$\text{CASE-III}:$

For , ${\sigma }_1+{\sigma }_2<0$ we get, ${\sigma }_1<-{\sigma }_2$

Thus, Charge on the left plate is lesser compared to the charge on the right plate.

So, Electric field in the region $\text{I}$ is towards right and in the region $\text{III}$ is towards left.

$\therefore \text{C}\to \text{R}$

Hence, option (b) is the correct answer.