Answer
$\mathrm{C} \lt \mathrm{A}\lt \mathrm{B}$
Work Step by Step
Electric charges exert forces on one another along the line connecting them.
Like charges repel, opposite charges attract.
The magnitude of the force between two point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$ is
$F =k\displaystyle \frac{|q_{1}||q_{2}|}{r^{2}},\qquad (19- 5)$
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The other two charges exert forces on the third. Take right to be the + direction.
Forces experienced by
$A:\displaystyle \qquad F=+k\frac{q^{2}}{d^{2}}+k\frac{q^{2}}{(2d)^{2}}=\frac{5}{4}\frac{kq^{2}}{d^{2}}$
$B:\displaystyle \qquad F=-k\frac{q^{2}}{d^{2}}-k\frac{q^{2}}{d^{2}}=-2\frac{kq^{2}}{d^{2}}$
$C:\displaystyle \qquad F=-k\frac{q^{2}}{(2d)^{2}}+k\frac{q^{2}}{d^{2}}=\frac{3}{4}\frac{kq^{2}}{d^{2}}$
Comparing magnitudes,
$\mathrm{C} \lt \mathrm{A}\lt \mathrm{B}$
