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)

Two electrons are fixed at positions $$(-d, 0)$$ and $$(d, 0)$$. A proton is placed at the midpoint (origin) and displaced slightly by a distance $$x$$ perpendicular to the line joining the electrons, where $$x \ll d$$.

The distance from the displaced proton to each electron is $$r = \sqrt{d^2 + x^2}$$. The attractive Coulomb force on the proton due to each electron has magnitude $$F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{d^2 + x^2}$$, where $$q$$ is the elementary charge.

By symmetry, the components of force along the line joining the electrons cancel out. The net restoring force is along the perpendicular direction. The perpendicular component from each electron is $$F\sin\theta = F\frac{x}{\sqrt{d^2 + x^2}}$$. The total restoring force is $$F_{\text{net}} = 2 \times \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(d^2 + x^2)} \times \frac{x}{\sqrt{d^2 + x^2}} = \frac{q^2 x}{2\pi\varepsilon_0(d^2 + x^2)^{3/2}}$$.

Since $$x \ll d$$, we approximate $$(d^2 + x^2)^{3/2} \approx d^3$$. Therefore, $$F_{\text{net}} \approx \frac{q^2}{2\pi\varepsilon_0 d^3}x$$.

This is a restoring force proportional to $$x$$, confirming SHM. Comparing with $$F = m\omega^2 x$$, we get $$\omega^2 = \frac{q^2}{2\pi\varepsilon_0 m d^3}$$.

Therefore, the angular frequency is $$\omega = \left(\frac{q^2}{2\pi\varepsilon_0 m d^3}\right)^{\frac{1}{2}}$$.

The correct answer is $$\left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{\frac{1}{2}}$$.