Two electrons each are fixed at a distance '2d'. A third charge… | JEE Main 2021 (24 Feb, Shift 2)

Q29JEE Main 2021 · 24 FEB · SHIFT 2Medium

Motion of Charges in an Electric Field

Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance x (x << d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: (m = mass of charged particle)

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Answer: (b)

Each electron attracts the proton with $F = \frac{kq^2}{d^2 + x^2}$ ; the components along the line cancel.
Net restoring force: $2F\frac{x}{\sqrt{d^2 + x^2}} \approx \frac{2kq^2}{d^3}x$ for $x \ll d$ .
$\omega^2 = \frac{2kq^2}{md^3} = \frac{q^2}{2\pi\varepsilon_0 m d^3}$
So the answer is (b) $\left(\frac{q^2}{2\pi\varepsilon_0 m d^3}\right)^{1/2}$ .

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Motion of Charges in an Electric Field question from JEE Main 2021 (24 Feb, Shift 2)

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