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(B) Let $a$ be the side length of the equilateral triangle and $r$ be the distance of each charge from the origin (circumradius).For an equilateral tr

Vedclass · Accepted Answer

$\left[\frac{\sqrt{3}}{12 \pi \varepsilon_0} \frac{q^2}{k}\right]^{1 / 3}$

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(B) Let $a$ be the side length of the equilateral triangle and $r$ be the distance of each charge from the origin (circumradius).For an equilateral triangle,$r = \frac{a}{\sqrt{3}}$,so $a = \sqrt{3} r$.The force on one charge due to the other two charges is the vector sum of the two Coulomb forces. Each force has magnitude $F_C = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2}$.The angle between these two forces is $60^\circ$. The net force $F_{	ext{net}}$ is directed towards the origin:$F_{	ext{net}} = \sqrt{F_C^2 + F_C^2 + 2 F_C^2 \cos 60^\circ} = \sqrt{3} F_C = \sqrt{3} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2} ight)$.Substituting $a^2 = 3 r^2$:$F_{	ext{net}} = \sqrt{3} \left( \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{3 r^2} ight) = \frac{q^2}{4 \pi \varepsilon_0 \sqrt{3} r^2}$.This force is balanced by the restoring force $F(r) = k r$:$k r = \frac{q^2}{4 \pi \varepsilon_0 \sqrt{3} r^2} \Rightarrow r^3 = \frac{q^2}{4 \pi \varepsilon_0 \sqrt{3} k} = \frac{\sqrt{3} q^2}{12 \pi \varepsilon_0 k}$.Thus,$r = \left[ \frac{\sqrt{3}}{12 \pi \varepsilon_0} \frac{q^2}{k} ight]^{1/3}$.

Three equal charges $+q$ are placed at the three vertices of an equilateral triangle centered at the origin. They are held in equilibrium by a restoring force of magnitude $F(r) = k r$ directed towards the origin,where $k$ is a constant. What is the distance of the three charges from the origin?

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