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
- Aπ ε 0 m d 3 2 q 2 1 2
- B2 π ε 0 m d 3 q 2 1 2
- Cq 2 2 π ε 0 m d 3 1 2
- 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