Question

Three point charges are placed at the vertices of a square of diagonal $4\text{m}$ as shown in figure.

What will be the net force experienced by a charge $-q$ placed at centre of a square ?

• A. $\frac{kQq}{4}(\hat{i}+\hat{j})$
• B. $\frac{kQq}{4}(\hat{i}-\hat{j})$
• C. $\frac{kQq}{4\sqrt{2}}(\hat{i}-\hat{j})$
• D. $\frac{kQq}{4\sqrt{2}}(\hat{i}+\hat{j})$

Answer 1

The correct option is D $\frac{kQq}{4\sqrt{2}}(\hat{i}+\hat{j})$

The force experienced by the charge $-q$ placed at the centre of the square is shown in the figure.
$\because$ Length of diagonal $=4\text{m}$.
$\therefore$ Length of $ao=\frac{4}{2}=2\text{m}$.

From above figure,
$F_{03}$ and $F_{01}$ have opposite direction and equal in magnitude, which is equal to

$|{\overrightarrow{F}}_{01}|=|{\overrightarrow{F}}_{02}|=|{\overrightarrow{F}}_{03}|=\frac{k\left(Q\right)\left(q\right)}{2^2}=\frac{kQq}{4}$

$\therefore$ Net effect of ${\overrightarrow{F}}_{01}$ and ${\overrightarrow{F}}_{03}$ on charge $-q$ is zero.

So, Net force experience by $-q$ is ${\overrightarrow{F}}_{02}$.
$\therefore {\overrightarrow{F}}_{02}=|{\overrightarrow{F}}_{02}|\cos {45}^{\circ }\hat{i}+|{\overrightarrow{F}}_{02}|\sin {45}^{\circ }\hat{j}$

$\Rightarrow {\overrightarrow{F}}_{02}=\frac{kQq}{4}.\frac{1}{\sqrt{2}}\hat{i}+\frac{kQq}{4}.\frac{1}{\sqrt{2}}\hat{j}$

$\Rightarrow {\overrightarrow{F}}_{02}=\frac{kQq}{4\sqrt{2}}(\hat{i}+\hat{j})$

Hence, option $\left(d\right)$ is the correct answer.