Question

Three particles, each of mass $m$ and carrying a charge $q$ are suspended from a common point by insulating massless strings, each of length $L$. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side $a$, then calculate the charge $q$ on each particle.
[Assume $L>>a$].

• A. ${\left[\frac{2\pi {ϵ}_0{a}^{3}mg}{3L}\right]}^{\frac{1}{2}}$
• B. None
• C. ${\left[\frac{4\pi {ϵ}_0{a}^{3}mg}{3L}\right]}^{\frac{1}{3}}$
• D. ${\left[\frac{4\pi {ϵ}_0{a}^{3}mg}{3L}\right]}^{\frac{1}{2}}$

Answer 1

The correct option is D ${\left[\frac{4\pi {ϵ}_0{a}^{3}mg}{3L}\right]}^{\frac{1}{2}}$

From Fig. (a), for equilibrium of a particle along a vertical line, we get

$T\cos \theta =mg\dots \dots \left(1\right)$

While for equilibrium in the plane of equilateral triangle, we get

$T\sin \theta =2F\cos {30}^{\circ }\dots \dots \left(2\right)$

So, from eqns. $\left(1\right)$ and $\left(2\right)$, we have

$\tan \theta =\frac{\sqrt{3}F}{mg}\dots \dots \left(3\right)$


Here, $F=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{a^2}$ and $\tan \theta =\frac{OA}{OP}=\frac{OA}{\sqrt{L^2-O{A}^2}}$

Also, from Fig. (c), we get

$OA=\frac{2}{3}AD=\frac{2}{3}a\sin {60}^{\circ }=\frac{a}{\sqrt{3}}$

So,

$\tan \theta =\frac{\left(\frac{a}{\sqrt{3}}\right)}{\sqrt{L^2-\left(\frac{a^2}{3}\right)}}=\frac{a}{\left(\sqrt{3}\right)L}(\text{as}L>>a)$

On substituting the above values of $F$ and $\tan \theta$ in Eq. $\left(3\right)$, we get

$\frac{a}{\left(\sqrt{3}\right)L}=\frac{\sqrt{3}}{mg}\frac{q^2}{4\pi {ϵ}_0{a}^2}$

$\Rightarrow q={\left[\frac{4\pi {ϵ}_0{a}^{3}mg}{3L}\right]}^{1/2}$

Hence, option (b) is the correct answer.