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Two charges $\mathrm{q}$ and $-3\mathrm{q}$ are placed fixed on $x-$ axis separated by distance $\mathrm{'d'}$. Where should a third charge $2\mathrm{q}$ be placed such that it will not experience any force ?
Solution
Suppose distance $\mathrm{AC}=x$ and $\mathrm{AB}=d$
Repulsive force on $2 q$ by $q$,
$\mathrm{F}_{q}=\frac{k(q)(2 q)}{x^{2}} \quad \ldots(1)$
Attractive force on $2 q$ by $-3 q$,
F-3q $=-\frac{k(2 q)(3 q)}{(x+d)^{2}} \ldots$ (2)
Resultant force on $2 q$,
F $=$ F $q+$ F $_{-3 q}$
O $=$ F $q+\mathrm{F}_{-3 q} \quad(\because$ No force on $2 q$
$\therefore$ Fq $=-\mathrm{F}_{-3 q}$
$\frac{k\left(2 q^{2}\right)}{x^{2}}=+\frac{k\left(6 q^{2}\right)}{(x+d)^{2}}$
$\therefore \frac{1}{x^{2}}=\frac{3}{(x+d)^{2}}$
$x^{2}+2 x d+d^{2}=3 x^{2}$
$\therefore 2 x d+d^{2}=2 x^{2}$
$\therefore 2 x^{2}-2 x d+d^{2}=0$
$\therefore \Delta=b^{2}-4 a c=(-2 d)^{2}-4 \times 2 \times d^{2}$
$=4 d^{2}+8 d^{2}=12 d^{2}$ $\therefore \sqrt{\Delta}=2 d(\sqrt{3})$ $\therefore \quad x=\frac{-b \pm \sqrt{\Delta}}{2 a}=\frac{2 d \pm 2 d(\sqrt{3})}{2 \times 2}$ $=\frac{d \pm \sqrt{3} d}{2}$ $=\frac{d}{2}[1 \pm \sqrt{3}]$ $=\frac{d}{2}[1+\sqrt{3}] \text { (Negative is not possible) }$
