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(D) Let the three charges be $q$ at the vertices of an equilateral triangle. Consider one charge at a vertex. It experiences two repulsive forces from

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$\sqrt{3} \,F$

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(D) Let the three charges be $q$ at the vertices of an equilateral triangle. Consider one charge at a vertex. It experiences two repulsive forces from the other two charges, each of magnitude $F$. The angle between these two forces is $60^{\circ}$ (the interior angle of an equilateral triangle). The resultant force $R$ is given by the vector addition formula: $R = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos 	heta}$ Substituting $F_1 = F$, $F_2 = F$, and $	heta = 60^{\circ}$: $R = \sqrt{F^2 + F^2 + 2 F^2 \cos 60^{\circ}}$ Since $\cos 60^{\circ} = 0.5$: $R = \sqrt{2 F^2 + 2 F^2 (0.5)} = \sqrt{2 F^2 + F^2} = \sqrt{3 F^2} = \sqrt{3} F$.

Three charges, each of $+1 \mu C$, are placed at the corners of an equilateral triangle. If the repulsive force between any two charges is $F$, then the net force on any one charge will be:

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