Net force on free charged particle,
F=(d+x)2kq2−(d−x)2kq2
F=−kq2[(d2−x2)24dx]
a=−m4kq2d(d4x)
a=−(md34kq2)x
So, angular frequency
ω=md34kq2
ω=1×10−6×134×9×109×10
ω=6×108radsec−1
JEE Main 2021 — Physics Electromagnetism
A particle of mass 1mg and charge q is lying at the mid-point of two stationary particles kept at a distance 2m when each is carrying same charge q. If the free charged particle is displaced from its equilibrium position through distance x (x<<1m). The particle executes SHM. Its angular frequency of oscillation will be _______ ×105rads−1 (if q2=10C2)
Held on 25 Jul 2021 · Verified 6 Jul 2026.
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