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:

• A. ${\left(\frac{q^2}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{md}}^3}\right)}^{1/2}$
• B. ${\left(\frac{{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{md}}^3}{2q^2}\right)}^{1/2}$
• C. ${\left(\frac{2{\mathrm{\pi \epsilon }}_{\circ }\circ {\mathrm{md}}^3}{q^2}\right)}^{1/2}$
• D. ${\left(\frac{2q^2}{{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{md}}^3}\right)}^{1/2}$

Answer 1

The correct option is A

${\left(\frac{q^2}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{md}}^3}\right)}^{1/2}$

The explanation for the correct option:

A: ${\left(\frac{q^2}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{md}}^3}\right)}^{1/2}$

Step 1: Given data

Distance between electrons$=2d$

Step 2: Formula used

The restoring force can be computed using the following formula:

$F_r=2F_1\sin \left(\theta \right)$

Step 3: Compute the restoring force on the proton.

Consider the following figure:

From the figure, it is clear that $F_1=\frac{2k{q}^2}{\left(d^2+x^2\right)}$. So,

$F_r=\frac{2K{q}^{2}x}{{\left(d^2+x^2\right)}^{3/2}}$

If $x$ is very very greater than $d$ then ,

$$\begin{gathered} F_r=\frac{2K{q}^{2}x}{d^3} \\ {F}_r=\frac{q^{2}x}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{d}}^3} \\ {F}_r=Kx \\ K=\frac{q^2}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{d}}^3} \end{gathered}$$

Step 4: Compute the angular frequency

Substitute the known values in the formula $\omega =\sqrt{\frac{K}{m}}$,

$\omega ={\left(\frac{q^2}{2{\mathrm{\pi \epsilon }}_{\circ }{\mathrm{d}}^3\mathrm{m}}\right)}^{\frac{1}{2}}$

Hence, option A is the correct answer.