From the given condition, we have

Fnetq=−[2Fq/qcosθ]
Fnetq=−2⋅4πϵ01⋅(d2+x2)2q2⋅d2+x2x
=−2πϵ0q2(d2+x2)3/2x
For x<<d,
Fnetq=−2πϵ0d3q2x
∴a=−2πϵ0⋅md3q2x
Comparing with equation of SHM (a=−ω2x)
∴ω=2πϵ0md3q2
JEE Main 2021 — Physics Electromagnetism
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)
Held on 24 Feb 2021 · Verified 6 Jul 2026.
(2q2πϵ0md3)21
(q22πϵ0md3)21
(2πϵ0md3q2)21
(πϵ0md32q2)21
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