Two identical positive charges $$Q$$ each are fixed at a distance of $$2a$$ apart from each other. Another point charge $$q_0$$ with mass $$m$$ is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $$q_0$$ executes SHM. The time period of oscillation of charge $$q_0$$ will be

Two identical positive charges $$Q$$ are fixed at a distance $$2a$$ apart, and a negative charge $$q_0$$ (mass $$m$$) is placed at the midpoint. We wish to find the time period of small oscillations of $$q_0$$ along the line joining the charges.

We place the two fixed charges at $$x=-a$$ and $$x=+a$$, with the negative charge $$q_0$$ initially at the origin. Suppose it is displaced by a small distance $$x$$ (where $$x\ll a$$) toward the charge at $$+a$$.

Since $$q_0$$ is negative and both fixed charges are positive, it experiences attractive forces from each. The force due to the charge at $$+a$$ (pulling toward $$+a$$) is $$F_{\text{right}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a-x)^2},$$ while the force due to the charge at $$-a$$ (pulling toward $$-a$$) is $$F_{\text{left}}=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{(a+x)^2}.$$

Taking the $$+x$$ direction as positive, the net force on $$q_0$$ is $$F_{\text{net}}=\frac{q_0Q}{4\pi\epsilon_0}\Bigl[\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\Bigr].$$ Because the charge is closer to $$+a$$ when displaced, the pull from that side is stronger, while the farther charge at $$-a$$ pulls back toward the origin. For negative $$q_0$$, this difference produces a restoring force toward $$x=0$$.

For small $$x$$ ($$x\ll a$$), we expand by the binomial theorem: $$\frac{1}{(a-x)^2}=\frac{1}{a^2}\Bigl(1-\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1+\frac{2x}{a}\Bigr),$$ $$\frac{1}{(a+x)^2}=\frac{1}{a^2}\Bigl(1+\frac{x}{a}\Bigr)^{-2}\approx\frac{1}{a^2}\Bigl(1-\frac{2x}{a}\Bigr).$$

Subtracting these approximations gives $$\frac{1}{(a-x)^2}-\frac{1}{(a+x)^2}\approx\frac{1}{a^2}\Bigl(\frac{4x}{a}\Bigr)=\frac{4x}{a^3}.$$

Substituting back into the expression for the force yields the magnitude of the restoring force $$|F|=\frac{q_0Q}{4\pi\epsilon_0}\cdot\frac{4x}{a^3}=\frac{q_0Q}{\pi\epsilon_0a^3}\,x.$$ This is of the form $$F=-kx$$, where $$k=\frac{q_0Q}{\pi\epsilon_0a^3}.$$

From the above, the angular frequency satisfies $$\omega^2=\frac{k}{m}=\frac{q_0Q}{\pi\epsilon_0ma^3},$$ and the time period is $$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{\pi\epsilon_0ma^3}{q_0Q}}=\sqrt{\frac{4\pi^3\epsilon_0ma^3}{q_0Q}}\,. $$

The correct answer is Option A: $$\sqrt{\dfrac{4\pi^3 \epsilon_0 m a^3}{q_0 Q}}$$.