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?
(N/A) Let the third charge $2q$ be placed at a distance $x$ from the charge $q$ on the side away from $-3q$.
The repulsive force on $2q$ due to $q$ is:
$F_q = \frac{k(q)(2q)}{x^2} = \frac{2kq^2}{x^2}$
The attractive force on $2q$ due to $-3q$ is:
$F_{-3q} = \frac{k(3q)(2q)}{(x+d)^2} = \frac{6kq^2}{(x+d)^2}$
For the net force to be zero,the magnitudes of these forces must be equal:
$F_q = F_{-3q}$
$\frac{2kq^2}{x^2} = \frac{6kq^2}{(x+d)^2}$
$\frac{1}{x^2} = \frac{3}{(x+d)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{\sqrt{3}}{x+d}$
$x+d = \sqrt{3}x$
$d = x(\sqrt{3}-1)$
$x = \frac{d}{\sqrt{3}-1} = \frac{d(\sqrt{3}+1)}{3-1} = \frac{d(\sqrt{3}+1)}{2}$
Thus,the charge $2q$ should be placed at a distance $\frac{d(\sqrt{3}+1)}{2}$ from the charge $q$ on the side away from $-3q$.
The repulsive force on $2q$ due to $q$ is:
$F_q = \frac{k(q)(2q)}{x^2} = \frac{2kq^2}{x^2}$
The attractive force on $2q$ due to $-3q$ is:
$F_{-3q} = \frac{k(3q)(2q)}{(x+d)^2} = \frac{6kq^2}{(x+d)^2}$
For the net force to be zero,the magnitudes of these forces must be equal:
$F_q = F_{-3q}$
$\frac{2kq^2}{x^2} = \frac{6kq^2}{(x+d)^2}$
$\frac{1}{x^2} = \frac{3}{(x+d)^2}$
Taking the square root on both sides:
$\frac{1}{x} = \frac{\sqrt{3}}{x+d}$
$x+d = \sqrt{3}x$
$d = x(\sqrt{3}-1)$
$x = \frac{d}{\sqrt{3}-1} = \frac{d(\sqrt{3}+1)}{3-1} = \frac{d(\sqrt{3}+1)}{2}$
Thus,the charge $2q$ should be placed at a distance $\frac{d(\sqrt{3}+1)}{2}$ from the charge $q$ on the side away from $-3q$.
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