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(C) Let the unit negative charge be at a distance $x$ from the midpoint along the perpendicular bisector. The distance of this charge from each fixed

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None of the above

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(C) Let the unit negative charge be at a distance $x$ from the midpoint along the perpendicular bisector. The distance of this charge from each fixed charge $+Q$ is $r = \sqrt{x^2 + a^2}$.The electrostatic force exerted by each $+Q$ charge on the unit negative charge is $F = \frac{1}{4 \pi \varepsilon_0} \frac{Q \cdot 1}{r^2} = \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)}$.The components of these forces perpendicular to the line connecting the charges cancel out, while the components along the perpendicular bisector add up.The net restoring force is $F_{	ext{net}} = -2F \cos 	heta$, where $\cos 	heta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + a^2}}$.$F_{	ext{net}} = -2 \left( \frac{Q}{4 \pi \varepsilon_0 (x^2 + a^2)} ight) \left( \frac{x}{\sqrt{x^2 + a^2}} ight) = -\frac{2Qx}{4 \pi \varepsilon_0 (x^2 + a^2)^{3/2}}$.Since $x \ll a$, we can approximate $(x^2 + a^2)^{3/2} \approx (a^2)^{3/2} = a^3$.Thus, $F_{	ext{net}} \approx -\left( \frac{2Q}{4 \pi \varepsilon_0 a^3} ight) x = -\left( \frac{Q}{2 \pi \varepsilon_0 a^3} ight) x$.This is the equation of simple harmonic motion $F = -kx_{eff}$, where $k_{eff} = \frac{Q}{2 \pi \varepsilon_0 a^3}$.The frequency of oscillation is $f = \frac{1}{2 \pi} \sqrt{\frac{k_{eff}}{M}} = \frac{1}{2 \pi} \sqrt{\frac{Q}{2 \pi \varepsilon_0 M a^3}}$.Comparing this with the given options, none of the options match the calculated frequency.

$A$ unit negative charge with mass $M$ resides at the mid-point of the straight line of length $2a$ connecting 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:

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