Question

Four point charges are placed at the vertices of a square of side $\sqrt{2}\text{m}$ as shown in figure.

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

Here $k=\frac{1}{4\pi {ϵ}_o}$

• A. $4kQq$
• B. $2kQq$
• C. $2\sqrt{2}kQq$
• D. $4kQq$

Answer 1

The correct option is C $2\sqrt{2}kQq$

The force experienced by the charge $+q$ placed at the centre of the square is shown in the figure.
Since, side $\left(a\right)$ of square is $\sqrt{2}\text{m}$.
From figure,
$AC=\sqrt{A{C}^2+B{C}^2}=\sqrt{2+2}=2\text{m}$

$\Rightarrow AO=\frac{AC}{2}=1\text{m}$

The distance between vortex and centre is $1\text{m}$.

So, Magnitude of force acting between $+q$ and each charge (at vertex) will be equal i.e. $F=F_{01}=F_{02}=F_{03}=F_{04}$.

So, from Coulomb's law

$|\overrightarrow{F}|=\frac{k\left(Q\right)\left(q\right)}{1^2}=kQq$

So, from given figure net force on charge $+q$ is $|{\overrightarrow{F}}_{net}|$

$\therefore |{\overrightarrow{F}}_{net}|=\sqrt{(2F{)}^2+(2F{)}^2}[\because \theta ={90}^{\circ }]$

$\Rightarrow |{\overrightarrow{F}}_{net}|=\sqrt{(2kQq{)}^2+(2kQq{)}^2}$

$\Rightarrow |{\overrightarrow{F}}_{net}|=2\sqrt{2}kQq$

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