Question

Two charges $q$ and $-3q$ are fixed on $x$-axis separated by distance $d$. Where should a third charge $2q$ be placed from $A$ such that it will not experience any force?

• A. $\frac{d-\sqrt{3}d}{2}$
• B. $\frac{d+\sqrt{3}d}{2}$
• C. $\frac{d+3d}{2}$
• D. $\frac{d-3d}{2}$

Answer 1

Answer: (B)

Let a charge $2q$ be placed at $P$, at a distance $l$ from $A$ where charge q is placed, as shown in figure.
The charge 2q will not experience any force, when force of repulsion on it due to $q$ is balanced by force of attraction on it due to $-3q$ at $B$ where $AB=d$
or $\frac{\left(2q\right)\left(q\right)}{4\pi {\epsilon }_0{l}^2}=\frac{\left(2q\right)\left(3q\right)}{4\pi {\epsilon }_0{\left(l+d\right)}^2}$
${\left(l+d\right)}^2=3l^2$ or $2l^2-2ld-d^2=0$
$\therefore l=\frac{2d\pm \sqrt{4d^2+8d^2}}{4}=\frac{d}{2}\pm \frac{\sqrt{3}d}{2}\Rightarrow l=\frac{d+\sqrt{3}d}{2}$
If charge $2q$ is placed between $A$ and $B$ then $l=\frac{\left(\sqrt{3}-1\right)}{2}d$