Question

Three equal negative charges, $-q_1$ each, form the vertices of an equilateral triangle. A particle of mass $m$ and a positive charge $q_2$ is constrained to move along a line perpendicular to the plane of triangle and through its centre which is at a distance $r$ from each of the negative charges as shown in the figure. The whole system is kept in gravity free space. Find the time period of vibration of the particle for small displacement from equilibrium position.

• A. $2\pi \sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{3q_1{q}_2}}$
• B. $2\pi \sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{q_1{q}_2}}$
• C. $2\pi \sqrt{\frac{3\pi {ϵ}_{0}m{r}^3}{4q_1{q}_2}}$
• D. $2\pi \sqrt{\frac{\pi {ϵ}_{0}m{r}^3}{3q_1{q}_2}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{3q_1{q}_2}}$

Let the charge $q_2$ is moved a distance $x$ perpendicular to the plane of the triangle.

Distance of charge $q_2$ from one of the vertices of triangle,
$R=\sqrt{r^2+x^2}$

Let the angle between $x$ and $R$ is $\theta$.
Net force on charge $q_2$, perpendicular to plane of triangle is $3F\cos \theta$

$F_{\text{net}}=-3F\cos \theta =-\frac{3k{q}_1{q}_2}{R^2}\frac{x}{R}=ma$
here $-$ve sign indicates that force is opposite to displacement.

$\Rightarrow a=-\frac{3k{q}_1{q}_{2}x}{m(x^2+r^2{)}^3}$

$\Rightarrow a=-\left[\frac{3q_1{q}_2}{4\pi {ϵ}_{0}m{r}^3}\right]x=-{\omega }^{2}x$ (as for small $x$ i.e. $r>>x$)

$\Rightarrow \omega =\sqrt{\frac{3q_1{q}_2}{4\pi {ϵ}_{0}m{r}^3}}$

$T=2\pi /\omega =2\pi \sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{3q_1{q}_2}}$
Hence, option (a) is correct.