# Two charges –q each are fixed separated by distance 2d. A third charge q of mass m placed at the mid-point is displaced slightly by x (x<<d) perpendicular to the line joining the two fixed charged as shown in Fig. 1.14. Show that q will perform simple harmonic oscillation of time period.

Let the charge q is displaced slightly by $x(x<perpendicular to the line joining the two fixed charges. The net force on the charge q will be towards O.The motion of charge -q to be simple harmonic if the force on charge q must be proportional to its distance from the centre O and is directed towards O.
The net force on the charge $F_{net}= 2F \cos\theta$
Here  $F=\frac{1}{4\pi \varepsilon _0}\frac{q(q)}{r^2}= \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )}$

And$\cos \theta =\frac{x}{\sqrt{x^2+d^2}}$

Hence, $F_{net}=2 \left [ \frac{1}{4\pi \varepsilon _0}\frac{q^2}{\left ( d^2+x^2 \right )} \right ]\left [ \frac{x}{\sqrt{x^2+d^2}} \right ]$

$= \frac{1}{2\pi \varepsilon _0}\frac{q^2x}{\left ( d^2+x^2 \right )^{3/2}}=\frac{1}{2\pi \varepsilon _0}\frac{q^2x}{d^3\left ( 1+\frac{x^2}{d^2} \right )^{3/2}}$

As $x<
i.e. force on charge q is proportional to its displacement from the centre O and it is directed towards O.

Hence, the motion of charge q would be simple harmonic, where

$\omega = \sqrt{\frac{K}{m}}\\ and \: T=\frac{2 \pi }{\omega}=2 \pi\sqrt{\frac{m}{K}}$
$\Rightarrow T= 2 \pi \sqrt{\frac{m.4 \pi \varepsilon _0d^3}{2q^2}}=\left [ \frac{8 \pi^3 \varepsilon _0md^3}{q^2} \right ]^{1/2}$

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