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(D) Consider the charge at vertex $A$. The charges at vertices $B$ and $C$ exert repulsive forces on the charge at $A$.Let $F_B$ be the force exerted

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$\frac{\sqrt{3} kQ^2}{a^2}$

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(D) Consider the charge at vertex $A$. The charges at vertices $B$ and $C$ exert repulsive forces on the charge at $A$.Let $F_B$ be the force exerted by the charge at $B$ on $A$,and $F_C$ be the force exerted by the charge at $C$ on $A$.According to Coulomb's law,the magnitude of these forces is $F_B = F_C = F = \frac{kQ^2}{a^2}$.The angle between these two forces is $60^\circ$ (the interior angle of an equilateral triangle).The net force $F_{net}$ on the charge at $A$ is given by the vector sum:$F_{net} = \sqrt{F_B^2 + F_C^2 + 2F_B F_C \cos 60^\circ}$Substituting $F_B = F_C = F$:$F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cos 60^\circ} = \sqrt{2F^2 + 2F^2(0.5)} = \sqrt{3F^2} = \sqrt{3}F$Substituting the value of $F$:$F_{net} = \frac{\sqrt{3} kQ^2}{a^2}$

Three charges,each $+Q$,are placed at the vertices of an equilateral triangle of side $a$. What is the net electrostatic force on any one charge? $\left( k = \frac{1}{4\pi \varepsilon _0} \right)$

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