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Two charges $-\mathrm{q}$ each are fixed separated by distance $2\mathrm{d}$. A third charge $\mathrm{d}$ of mass $m$ placed at the midpoint is displaced slightly by $x (x \,<\,<\, d)$ perpendicular to the line joining the two fixed charged as shown in figure. Show that $\mathrm{q}$ will perform simple harmonic oscillation of time period. $T =\left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$
Solution
Suppose charge at $\mathrm{A}$ and $\mathrm{B}$ are $-q$ and $\mathrm{O}$ is mid point of $\mathrm{AB}$ and $\mathrm{PO}$ is $x .$
$\begin{aligned} \therefore \mathrm{AB} &=\mathrm{AO}+\mathrm{OB} \\ &=d+d \\ &=2 d \end{aligned}$
$x$
$m$ is mass of charge $q .$
Attractive force by each charge $\mathrm{A}$ and $\mathrm{B}$ on charge at $\mathrm{P}$,
F $=\frac{k(q)(q)}{r^{2}}$
where $r=$ AP $=$ BP
Fsin $\theta$ components of force are of same magnitude but in opposite directions hence, their resultant
is zero and Fcos $\theta$ components are in same direction,
$\mathrm{F}^{\prime}=2 \mathrm{~F} \cos \theta$
$=\frac{2 k q^{2}}{r^{2}} \cos \theta$
But from figure, $r=\sqrt{d^{2}+x^{2}}$ and $\cos \theta=\frac{x}{r}$
$\therefore \mathrm{F}^{\prime}=\frac{2 k q^{2}}{\left(d^{2}+x^{2}\right)^{2}} \cdot \frac{x}{\left(d^{2}+x^{2}\right)^{1 / 2}}$
$=\frac{2 k q^{2} x}{\left(d^{2}+x^{2}\right)^{3 / 2}}$
