Question

The electric force on a charge $q_1$ due to $q_2$ is $(4\hat{i}-3\hat{j})\text{N}$. The unit vector in the direction of electirc force on $q_2$ due to $q_1$ will be:

• A. $-4\hat{i}+3\hat{j}$
• B. $\frac{-4}{5}\hat{i}+\frac{3}{5}\hat{j}$
• C. $-\frac{4}{5}\hat{i}-\frac{3}{5}\hat{j}$
• D. $\frac{4}{5}\hat{i}+\frac{3}{5}\hat{j}$

Answer 1

The correct option is B $\frac{-4}{5}\hat{i}+\frac{3}{5}\hat{j}$

From Newton's third law,
the interaction force between charges is equal and opposite in nature.
$\Rightarrow {\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21}$

$\because {\overrightarrow{F}}_{12}=(4\hat{i}-3\hat{j})\text{N}$

Thus, ${\overrightarrow{F}}_{21}=-{\overrightarrow{F}}_{12}$

$\Rightarrow {\overrightarrow{F}}_{21}=(-4\hat{i}+3\hat{j})\text{N}$

The unit vector in the direction of electric force on $q_2$ is,
${\overrightarrow{F}}_{21}=\frac{{\overrightarrow{F}}_{21}}{|{\overrightarrow{F}}_{21}|}$

$\Rightarrow {\overrightarrow{F}}_{21}=\frac{-4\hat{i}+3\hat{j}}{\sqrt{16+9}}$

$\therefore {\overrightarrow{F}}_{21}=\frac{-4}{5}\hat{i}+\frac{3}{5}\hat{j}$

Hence, option $\left(b\right)$ is correct.