A unit negative charge with mass \(M\) resides at the midpoint of the straight line of length \(2a\) adjoining two fixed charges of magnitude \(+Q\) each. If it is given a very small displacement \(x(x\ll a)\) in a direction perpendicular to the straight line, it will
execute oscillations with frequency \(\frac{1}{2\pi}\sqrt{\frac{2Q}{4\pi\varepsilon_0 Ma^3}}\)
Principle: if a small displacement produces a force directed back towards the equilibrium point and proportional to the displacement, \(F=-kx\), the body performs simple harmonic motion with \(\omega=\sqrt{k/M}\) and \(f=\frac{\omega}{2\pi}\).
Displaced by \(x\) perpendicular to the line, the unit negative charge is at distance \(r=\sqrt{a^2+x^2}\) from each \(+Q\). Each attraction has magnitude \(F_1=\frac{1}{4\pi\varepsilon_0}\frac{Q\cdot 1}{a^2+x^2}\), directed towards that charge.
Components along the line cancel; the perpendicular components, each \(F_1\frac{x}{r}\), add and point back towards the midpoint: \(F=-\frac{2Q}{4\pi\varepsilon_0}\frac{x}{(a^2+x^2)^{3/2}}\).
For \(x\ll a\), \((a^2+x^2)^{3/2}\approx a^3\), so \(F\approx-\frac{2Q}{4\pi\varepsilon_0 a^3}\,x\). The restoring force is linear in \(x\) with \(k=\frac{2Q}{4\pi\varepsilon_0 a^3}\).
So the charge neither stays at the midpoint (it overshoots and keeps moving) nor flies away; it oscillates with \(\omega=\sqrt{\frac{k}{M}}=\sqrt{\frac{2Q}{4\pi\varepsilon_0 Ma^3}}\), i.e. \(f=\frac{1}{2\pi}\sqrt{\frac{2Q}{4\pi\varepsilon_0 Ma^3}}\). The expression with \(a^2\) does not even have the dimension of a frequency.
Hence, the answer is execute oscillations with frequency \(\frac{1}{2\pi}\sqrt{\frac{2Q}{4\pi\varepsilon_0 Ma^3}}\).
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