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 $\frac{\left(2q\right)\left(q\right)}{4\pi\varepsilon_{0}l^{2}}=\frac{\left(2q\right)\left(3q\right)}{4\pi\varepsilon_{0}\left(l+d\right)^{2}}$
$\left(l+d\right)^{2}=3l^{2}$ 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 $l= \frac{\left(\sqrt{3}-1\right)}{2}d$