Import Question JSON

$\text{Hint: } a = \omega^2 x$ $\text{Step: Find the angular frequency by which the proton will undergo simple harmonic motion.}$ $\text{From the given condition, we have:}$ $\text{The forces on the proton due to the two electrons are equal in magnitude but opposite in direction along the line joining the electrons.}$ $\text{Only the perpendicular components of these forces contribute to the net restoring force.}$ $\text{The force due to a single electron at a distance } r \text{ from the proton is given by:}$ $F = \frac{q^2}{4 \pi \varepsilon_0 r^2}$ $\text{If the proton is displaced perpendicular to the line joining the electrons by a distance } x, \text{ the distance from the proton to each electron becomes:}$ $r = \sqrt{d^2 + x^2} \approx d \,(\text{since } x \ll d)$ $\text{The perpendicular component of the force due to one electron is given by:}$ $F_\perp = F \times \frac{x}{r} = \frac{q^2}{4 \pi \varepsilon_0 r^2} \times \frac{x}{r}$ $\text{Substituting } r \approx d \text{ we get:}$ $F_\perp = \frac{q^2}{4 \pi \varepsilon_0 d^2} \times \frac{x}{d} = \frac{q^2 x}{4 \pi \varepsilon_0 d^3}$ $\text{The net restoring force is doubled because both electrons contribute that is defined by:}$ $F_{\text{net}} = 2F_\perp = \frac{2q^2 x}{4 \pi \varepsilon_0 d^3}$ $\text{The equation of motion for the proton is given by:}$ $F_{\text{net}} = -kx$ $k = \frac{2q^2}{4 \pi \varepsilon_0 d^3}$ $\text{The angular frequency } \omega \text{ of simple harmonic motion is given by:}$ $\omega = \sqrt{\frac{k}{m}}$ $\omega = \sqrt{\frac{2q^2}{4 \pi \varepsilon_0 d^3 m}} = \sqrt{\frac{q^2}{2 \pi \varepsilon_0 m d^3}}$ $\text{Hence, option (3) is the correct answer.}$