A particle of mass 1 mg and charge $$q$$ is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge $$q$$. If the free charged particle is displaced from its equilibrium position through distance $$x$$ ($$x << 1$$ m). The particle executes SHM. Its angular frequency of oscillation will be ___ $$\times 10^5$$ rad s$$^{-1}$$ (if $$q^2 = 10C^2$$)

Let us first mark the fixed, identical charges. They are kept 2 m apart, so each is 1 m away from the mid-point. The free particle of charge $$q$$ and mass $$m$$ (here $$m = 1\text{ mg} = 1\times10^{-6}\,\text{kg}$$) is initially at this mid-point, where the forces cancel out.

Suppose the free particle is displaced very slightly ($$x\ll1\text{ m}$$) along the line joining the fixed charges. Choose the positive x-axis from the left charge towards the right charge. After a displacement $$x$$ to the right, its coordinates and distances become:

$$ \begin{aligned} \text{Position of particle}&:& +x,\\[4pt] \text{Distance to right charge}&:& r_R = (1 - x),\\[4pt] \text{Distance to left charge}&:& r_L = (1 + x). \end{aligned} $$

The electrostatic force between two like charges is given by Coulomb’s law

$$ F = k\frac{q^2}{r^2},\qquad k = \frac1{4\pi\varepsilon_0}=9\times10^{9}\,\text{N\,m}^2\text{C}^{-2}. $$

Because the charges are like, each fixed charge repels the free charge. Hence:

$$ \begin{aligned} \text{Force by right charge}&:& \vec F_R = -\,k\frac{q^{2}}{(1-x)^{2}}\hat i,\\[6pt] \text{Force by left charge}&:& \vec F_L = +\,k\frac{q^{2}}{(1+x)^{2}}\hat i. \end{aligned} $$ (The negative sign on $$\vec F_R$$ shows that it acts towards the left, while the positive sign on $$\vec F_L$$ shows that it acts towards the right.)

The net force on the particle is therefore

$$ \vec F = \vec F_L + \vec F_R = kq^{2}\left[\frac1{(1+x)^{2}} - \frac1{(1-x)^{2}}\right]\hat i. $$

For very small $$x$$ we may expand each reciprocal square by the binomial approximation $$\displaystyle(1\pm x)^{-2}\approx 1\mp2x+3x^{2}\ldots$$ Retaining only the first-order term in $$x$$ gives

$$ \begin{aligned} \frac1{(1+x)^{2}} &\approx 1 - 2x,\\ \frac1{(1-x)^{2}} &\approx 1 + 2x. \end{aligned} $$

Substituting these into the expression for $$\vec F$$ we obtain

$$ \vec F \approx kq^{2}\big[(1-2x) - (1+2x)\big]\hat i = kq^{2}(-4x)\hat i = -4kq^{2}x\,\hat i. $$

This force is proportional to $$-x$$, i.e. directed opposite to the displacement, which is the hallmark of simple harmonic motion (SHM). Comparing with the standard SHM equation

$$ m\frac{d^{2}x}{dt^{2}} = -\omega^{2}x, $$

we identify

$$ \omega^{2} = \frac{4kq^{2}}{m},\qquad\text{so}\qquad \omega = 2\sqrt{\frac{kq^{2}}{m}}. $$

Now we substitute the numerical values. We are given $$q^{2}=10\,\text{C}^{2}$$, hence we need only $$q^{2}$$, not $$q$$ itself.

$$ \begin{aligned} \omega &= 2\sqrt{\frac{(9\times10^{9})\,(10)}{1\times10^{-6}}}\\[6pt] &= 2\sqrt{9\times10^{10+6}}\\[6pt] &= 2\sqrt{9\times10^{16}}\\[6pt] &= 2\,(3\times10^{8})\\[6pt] &= 6\times10^{8}\ \text{rad s}^{-1}. \end{aligned} $$

The problem asks for the coefficient in

$$ \omega = (\,\text{blank}\,)\times10^{5}\ \text{rad s}^{-1}. $$

Dividing our result by $$10^{5}$$ gives

$$ \frac{6\times10^{8}}{10^{5}} = 6\times10^{3} = 6000. $$

So, the answer is $$6000$$.