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

Motion of Charges in an Electric Field question from 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)

  1. Each electron attracts the proton with F=kq2d2+x2F = \frac{kq^2}{d^2 + x^2}; the components along the line cancel.
  2. Net restoring force: 2Fxd2+x2≈2kq2d3x2F\frac{x}{\sqrt{d^2 + x^2}} \approx \frac{2kq^2}{d^3}x for x≪dx \ll d.
  3. ω2=2kq2md3=q22πε0md3\omega^2 = \frac{2kq^2}{md^3} = \frac{q^2}{2\pi\varepsilon_0 m d^3}
  4. So the answer is (b) (q22πε0md3)1/2\left(\frac{q^2}{2\pi\varepsilon_0 m d^3}\right)^{1/2}.
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