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. ${\left(\frac{2q^2}{\pi {\epsilon }_{0}m{d}^3}\right)}^{\frac{1}{2}}$
• B. ${\left(\frac{\pi {\epsilon }_{0}m{d}^3}{2q^2}\right)}^{\frac{1}{2}}$
• C. ${\left(\frac{q^2}{2\pi {\epsilon }_{0}m{d}^3}\right)}^{\frac{1}{2}}$
• D. ${\left(\frac{2\pi {\epsilon }_{0}m{d}^3}{q^2}\right)}^{\frac{1}{2}}$

Answer 1

Answer: (C)

From the given condition, we have

$F_{\text{netq }}=-\left[2F_{q/q}\cos \theta \right]$
$F_{\text{netq }}=-2\cdot \frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{{\left(\sqrt{d^2+x^2}\right)}^2}\cdot \frac{x}{\sqrt{d^2+x^2}}$
$=-\frac{q^2}{2\pi {\epsilon }_0}\frac{x}{{\left(d^2+x^2\right)}^{3/2}}$
For $x<<d$
$F_{netq}=-\frac{q^2}{2\pi {\epsilon }_0{d}^3}x$
$\therefore a=-\frac{q^2}{2\pi {\epsilon }_0\cdot m{d}^3}x$
Comparing with equation of $SHM\left(a=-{\omega }^{2}x\right)$
$\therefore \omega =\sqrt{\frac{q^2}{2\pi {\epsilon }_{0}m{d}^3}}$