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. $\left[\frac{1}{6\pi\varepsilon_{0}} \frac{q^{2}}{k}\right]^{^{^{1 /2}}}$
• B. $\left[\frac{\sqrt{3}}{12\pi\varepsilon_{0}} \frac{q^{2}}{k}\right]^{^{^{1 /3}}}$
• C. $\left[\frac{1}{6\pi\varepsilon_{0}} \frac{q^{2}}{k}\right]^{^{^{2/3}}}$
• D. $\left[\frac{\sqrt{3}}{4\pi\varepsilon_{0}} \frac{q^{2}}{k}\right]^{^{^{2/ 3}}}$

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}} $
$=\frac{\sqrt{3 }.q^{2}}{4\pi \varepsilon _{0}a^{2}}$
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, $F_{\text{net}}= \frac{\sqrt{3}.q^{2}}{\left(4\pi\varepsilon_{0}\right)\sqrt{3}r^{2}}$
or$F_{\text {net }}=\frac{q^{2}}{\left(4 \pi \varepsilon_{0}\right) \sqrt{3} r^{2}}$
Now, given that this force is balanced bya force $Fr\left(r\right) = kr$
$\therefore kr =\frac{q^{2}}{\left(4\pi\varepsilon_{0}\right)\sqrt{3}r^{2}}$
$\Rightarrow r^{3}=\frac{\sqrt{3}q^{2}}{12\pi\varepsilon_{0}k}$
So, $r=\left(\frac{\sqrt{3}q^{2}}{12\pi\varepsilon_{0}k}\right)^{^{\frac{1}{3}}}$