1. Set up the configuration

Two fixed particles, each carrying charge $q$, are placed symmetrically at distance $2\,\text{m}$ apart. So the midpoint is $1\,\text{m}$ from each fixed charge.

A third particle of mass $m=1\,\text{mg}$ and charge $q$ is placed at the midpoint.

Given: $m=1\,\text{mg}=10^{-6}\,\text{kg}, \qquad q^2=10\,\text{C}^2$

Let the middle particle be displaced by a small distance $x$ along the line joining the two fixed charges, where $x\ll 1\,\text{m}$.

2. Write the electrostatic forces

After displacement, its distances from the two fixed charges become: $1-x \quad \text{and} \quad 1+x$

Since all charges are same, the forces are repulsive.

• Force due to left charge (towards right): $F_1 = k\frac{q^2}{(1-x)^2}$

• Force due to right charge (towards left): $F_2 = k\frac{q^2}{(1+x)^2}$

Net force on the particle: $F = kq^2\left[\frac{1}{(1-x)^2}-\frac{1}{(1+x)^2}\right]$ But this force is away from the midpoint for same-sign charges, so equilibrium is unstable along the line. For SHM, the intended small oscillation must be for displacement perpendicular to the line joining the charges. So we now analyze the physically correct SHM direction.

3. Displacement perpendicular to the line joining charges

Let the particle be displaced by a small distance $x$ perpendicular to the line of centers.

Distance from each fixed charge becomes: $r=\sqrt{1+x^2}$

Force due to each charge: $F_0 = k\frac{q^2}{r^2}=k\frac{q^2}{1+x^2}$

Its vertical component is: $F_{0y}=F_0\frac{x}{r}=k\frac{q^2 x}{(1+x^2)^{3/2}}$

Both charges contribute equally, so total force is: $F_y = 2k\frac{q^2 x}{(1+x^2)^{3/2}}$ This is directed away from the midpoint for same-sign charges, again unstable.

Thus, for SHM to occur, the middle charge must be of opposite sign to the fixed charges (standard intended interpretation), giving restoring force: $F_y = -2k\frac{q^2 x}{(1+x^2)^{3/2}}$

For small $x$, $(1+x^2)^{-3/2} \approx 1$ So, $F_y \approx -2kq^2 x$

Comparing with SHM form: $F=-m\omega^2 x$ we get $m\omega^2 = 2kq^2$

Hence, $\omega = \sqrt{\frac{2kq^2}{m}}$

4. Substitute values

Using: $k=9\times 10^9\,\text{N m}^2/\text{C}^2, \quad q^2=10^{-10}\,\text{C}^2, \quad m=10^{-6}\,\text{kg}$

Then, $\omega = \sqrt{\frac{2\times 9\times 10^9\times 10^{-10}}{10^{-6}}}$

$\omega = \sqrt{\frac{18\times 10^{-1}}{10^{-6}}}$

$\omega = \sqrt{1.8\times 10^6}$

$\omega \approx 1.34\times 10^3\,\text{rad/s}$

This does not match the answer format unless the printed quantity is interpreted as: $q^2=10^{-8}\,\text{C}^2$

Then, $\omega = \sqrt{\frac{2\times 9\times 10^9\times 10^{-8}}{10^{-6}}}$ = \sqrt{18\times 10^7} \approx 4.24\times 10^4,\text{rad/s}$$

Still not matching.

If instead the intended motion is along the line and opposite-sign charges are assumed, then $F = kq^2\left[\frac{1}{(1+x)^2}-\frac{1}{(1-x)^2}\right]$ For small $x$, $\frac{1}{(1+x)^2}\approx 1-2x, \qquad \frac{1}{(1-x)^2}\approx 1+2x$

So, $F \approx kq^2[(1-2x)-(1+2x)] = -4kq^2x$

Thus, $m\omega^2 = 4kq^2$ $\omega = \sqrt{\frac{4kq^2}{m}}$

Now using the likely intended value $q^2=10^{-10}\,\text{C}^2$, $\omega = \sqrt{\frac{4\times 9\times 10^9\times 10^{-10}}{10^{-6}}}$ = \sqrt{3.6\times 10^6} \approx 1.9\times 10^3,\text{rad/s}$$

Again not matching 6000.

5. Conclusion

The given statement has an internal inconsistency:

• with all three charges same sign, midpoint equilibrium is not stable in the oscillation directions considered,
• the provided numerical data/answer do not fit consistently.

The closest standard textbook result for small oscillation along the line with restoring force is: $\omega = \sqrt{\frac{4kq^2}{m}}$

Using the stored answer format "$\_\_\_\_ \times 10^5$ rad/s" and stored answer $6000$, the implied angular frequency is: $\omega = 6000\times 10^{-5} = 0.06\,\text{rad/s}$ or possibly $6000\times 10^5$ depending on formatting, both of which are inconsistent with the physics/data.

Hence I do not agree with the stored correct answer.