Two electrons each are fixed at a distance $2d$. A third charge proton placed at the midpoint is displaced slightly by a distance $x\ (x \ll d)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
($m$ = mass of charged particle)
Answer: (C) $\left(\frac{q^2}{2\pi\varepsilon_0md^3}\right)^{\frac12}$
Each electron attracts the proton with force $\dfrac{q^2}{4\pi\varepsilon_0(d^2 + x^2)}$. Components along the line of the charges cancel; the components towards the midpoint add:
$$F = 2\cdot\frac{q^2}{4\pi\varepsilon_0(d^2+x^2)}\cdot\frac{x}{\sqrt{d^2+x^2}} \approx \frac{q^2x}{2\pi\varepsilon_0d^3}\quad(x \ll d)$$
This is a restoring force $F = -m\omega^2x$:
$$\omega = \left(\frac{q^2}{2\pi\varepsilon_0md^3}\right)^{1/2}$$
Solution by Sreeraj P, M.Sc Physics