Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.

(N/A) Let the charges at $A$ and $B$ be $-q$ and $O$ be the midpoint of $AB$. The charge $q$ is at point $P$ such that $PO = x$.
The distance from each fixed charge to $P$ is $r = \sqrt{d^2 + x^2}$.
The electrostatic force exerted by each charge $-q$ on charge $q$ is $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.
The components of force perpendicular to $PO$ cancel out,while the components along $PO$ add up:
$F_{net} = 2F \cos \theta = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} \right) \frac{x}{r} = \frac{2q^2 x}{4\pi\epsilon_0 (d^2 + x^2)^{3/2}}$.
Since $x \ll d$,we can approximate $(d^2 + x^2)^{3/2} \approx (d^2)^{3/2} = d^3$.
Thus,$F_{net} = \frac{2q^2 x}{4\pi\epsilon_0 d^3} = \frac{q^2}{2\pi\epsilon_0 d^3} x$.
Since the force is directed towards the equilibrium position $O$ and is proportional to displacement $x$,the motion is simple harmonic with restoring force $F = -kx_{eff}$,where $k = \frac{q^2}{2\pi\epsilon_0 d^3}$.
The time period is $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m \cdot 2\pi\epsilon_0 d^3}{q^2}} = \sqrt{\frac{4\pi^2 \cdot 2\pi\epsilon_0 m d^3}{q^2}} = \left[\frac{8\pi^3 \epsilon_0 m d^3}{q^2}\right]^{1/2}$.