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)

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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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