Question

Two electrons having charge $(-e)$ each are fixed at a distance $\text{‘2d’}$. A third charge proton placed at the midpoint is displaced slightly by a distance $\text{x (x<<d)}$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
$(\text{m}=$ mass of charged particle)

• A. ${\left(\frac{e^2}{2\pi {\epsilon }_{o}m{d}^3}\right)}^{1/2}$
• B. ${\left(\frac{\pi {\epsilon }_{o}m{d}^3}{2e^2}\right)}^{1/2}$
• C. ${\left(\frac{2\pi {\epsilon }_{o}m{d}^3}{e^2}\right)}^{1/2}$
• D. ${\left(\frac{2e^2}{\pi {\epsilon }_{o}m{d}^3}\right)}^{1/2}$

Answer 1

The correct option is A ${\left(\frac{e^2}{2\pi {\epsilon }_{o}m{d}^3}\right)}^{1/2}$


Restoring force on proton:-

$F_{\text{r}}=2F_1\sin \theta$, Where, $F_1=\frac{-k{e}^2}{d^2+x^2}$

$F_{\text{r}}=\frac{-2k{e}^{2}x}{(d^2+x^2{)}^{3/2}}$

$\because x<<d$

$F_{\text{r}}=\frac{-2k{e}^{2}x}{d^3}=-Kx$

$K=\frac{2k{e}^2}{d^3}=\frac{e^2}{2\pi {\epsilon }_o{d}^3}$

Angular Frequency :-

$\omega =\sqrt{\frac{K}{m}}$

$\therefore \omega =\sqrt{\frac{e^2}{2\pi {\epsilon }_{o}m{d}^3}}$