Question

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

• A. ${\left[\frac{1}{6\pi {\epsilon }_0}\frac{q^2}{k}\right]}^{{}^{{}^{1/2}}}$
• B. ${\left[\frac{\sqrt{3}}{12\pi {\epsilon }_0}\frac{q^2}{k}\right]}^{{}^{{}^{1/3}}}$
• C. ${\left[\frac{1}{6\pi {\epsilon }_0}\frac{q^2}{k}\right]}^{{}^{{}^{2/3}}}$
• D. ${\left[\frac{\sqrt{3}}{4\pi {\epsilon }_0}\frac{q^2}{k}\right]}^{{}^{{}^{2/3}}}$

Answer 1

Answer: (B)

In given charge configuration

Net force on any of charge is
$F_{\text{net}}=\sqrt{F_A^2+F_B^2+2F_A{F}_B\cos 60^{\circ }}$
$=\frac{\sqrt{3}.q^2}{4\pi {\epsilon }_0{a}^2}$
where, $a=$side length of equilateraltriangle
So, radius $r$ is $r$ $\frac{2}{3}\left(\frac{\sqrt{3}}{2}a\right)$
$\Rightarrow a=\sqrt{3}r$
Hence, $F_{\text{net}}=\frac{\sqrt{3}.q^2}{\left(4\pi {\epsilon }_0\right)\sqrt{3}{r}^2}$
or$F_{\text{net }}=\frac{q^2}{\left(4\pi {\epsilon }_0\right)\sqrt{3}{r}^2}$
Now, given that this force is balanced bya force $Fr\left(r\right)=kr$
$\therefore kr=\frac{q^2}{\left(4\pi {\epsilon }_0\right)\sqrt{3}{r}^2}$
$\Rightarrow r^3=\frac{\sqrt{3}{q}^2}{12\pi {\epsilon }_{0}k}$
So, $r={\left(\frac{\sqrt{3}{q}^2}{12\pi {\epsilon }_{0}k}\right)}^{{}^{\frac{1}{3}}}$