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(A) Let the charge $2q$ be placed at a distance $x$ from the charge $q$ on the side of $q$ (outside the region between the two charges). For the net f

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$\frac{d}{2}(1 + \sqrt{3})$ from $q$

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(A) Let the charge $2q$ be placed at a distance $x$ from the charge $q$ on the side of $q$ (outside the region between the two charges). For the net force on $2q$ to be zero,the magnitude of the force due to $q$ must equal the magnitude of the force due to $-3q$. $\frac{k(q)(2q)}{x^2} = \frac{k(3q)(2q)}{(d+x)^2}$ $\frac{1}{x^2} = \frac{3}{(d+x)^2}$ $(d+x)^2 = 3x^2$ $d+x = \pm \sqrt{3}x$ Since $x$ must be positive,we take $d+x = \sqrt{3}x$. $d = x(\sqrt{3} - 1)$ $x = \frac{d}{\sqrt{3}-1} = \frac{d(\sqrt{3}+1)}{3-1} = \frac{d}{2}(1+\sqrt{3})$ Thus,the charge $2q$ should be placed at a distance $\frac{d}{2}(1+\sqrt{3})$ from $q$ on the side away from $-3q$.

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

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