Question

A particle of mass $M$ and charge $q$ is at rest at the midpoint between two other fixed similar charges each of magnitude $Q$ placed at a distance $2d$ apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount $x(x<<d)$ along the line joining the two charges and is left to itself. It will now oscillate about the mean position with a time period :

(${ϵ}_0$ = permittivity of free space)

• A. $2\sqrt{\frac{{\pi }^{3}M{ϵ}_{0}d}{Qq}}$
• B. $2\sqrt{\frac{{\pi }^{2}M{ϵ}_0{d}^3}{Qq}}$
• C. $2\sqrt{\frac{{\pi }^{3}M{ϵ}_0{d}^3}{Qq}}$
• D. $2\sqrt{\frac{{\pi }^{3}M{ϵ}_0}{Qq{d}^3}}$

Answer 1

The correct option is C $2\sqrt{\frac{{\pi }^{3}M{ϵ}_0{d}^3}{Qq}}$

Net force acting on the charge $q$ $F=F_1-F_2$

$\therefore$ $F=\frac{qQ}{4\pi {ϵ}_o(d-x{)}^2}-\frac{qQ}{4\pi {ϵ}_o(d+x{)}^2}$

OR $F=\frac{qQ}{4\pi {ϵ}_o}[\frac{1}{(d-x{)}^2}-\frac{1}{(d+x{)}^2}]$

OR $F=\frac{qQ}{4\pi {ϵ}_o}[\frac{(d+x{)}^2-(d-x{)}^2}{(d^2-x^2)2}]$

$⟹$ $F=\frac{qQ}{4\pi {ϵ}_o}.\frac{4dx}{d^4(1-x^2/d^2{)}^2}=\frac{qQx}{\pi {ϵ}_o{d}^3(1-x^2/d^2{)}^2}$

For $x<<d$, we get $F=\frac{qQ}{\pi {ϵ}_o{d}^3}x=Kx$ where $K=\frac{Qq}{\pi {ϵ}_o{d}^3}$

Time period of oscillation $T=2\pi \sqrt{\frac{M}{K}}$

$\therefore$ $T=2\pi \sqrt{\frac{M}{qQ/\pi {ϵ}_o{d}^3}}=2\sqrt{\frac{{\pi }^{3}M{ϵ}_o{d}^3}{Qq}}$