Question

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 << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency : (m = mass of charged particle)

• A. $\left( \frac{q^2}{2\pi \varepsilon_0 md^3} \right)^{\frac{1}{2}}$
• B. $\left( \frac{2q^2}{\pi \varepsilon_0 md^3} \right)^{\frac{1}{2}}$
• C. $\left( \frac{\pi \varepsilon_0 md^3}{2q^2} \right)^{\frac{1}{2}}$
• D. $\left( \frac{2\pi \varepsilon_0 md^3}{q^2} \right)^{\frac{1}{2}}$

Hint:

For small displacements $x \ll d$, the binomial approximation or simply ignoring $x^2$ in $(d^2+x^2)$ terms is the key to proving SHM.

Answer 1

The Correct Option is A

Step 1: Restoring force $F = 2 F_e \cos\theta \approx 2 (\frac{1}{4\pi\epsilon_0} \frac{q^2}{d^2}) (\frac{x}{d})$.
Step 2: $F = \frac{q^2}{2\pi\epsilon_0 d^3} x$.
Step 3: In SHM, $F = m \omega^2 x$.
Step 4: $\omega^2 = \frac{q^2}{2\pi\epsilon_0 md^3} \Rightarrow \omega = \sqrt{\frac{q^2}{2\pi\epsilon_0 md^3}}$.