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JEE Main202124 Feb 2021Evening ShiftPhysicsElectrostaticsActual

Two electrons each are fixed at a distance 2 d . 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)

Options

  1. Aπ ε 0 m d 3 2 q 2 1 2
  2. B2 π ε 0 m d 3 q 2 1 2
  3. Cq 2 2 π ε 0 m d 3 1 2
  4. D2 q 2 π ε 0 m d 3 1 2

Correct answer

C. q 2 2 π ε 0 m d 3 1 2

Step-by-step solution

From the given condition, we have F net   q = - 2 F q / q cos θ F net   q = - 2 · 1 4 π ε 0 · q 2 d 2 + x 2 2 · x d 2 + x 2 = - q 2 2 π ε 0 x d 2 + x 2 3 / 2 For x < < d , F net   q = - q 2 2 π ε 0   d 3 x ∴     a = - q 2 2 π ε 0 · m d 3 x Comparing with equation of SHM a = - ω 2 x ∴     ω = q 2 2 π ε 0 m d 3

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