From conservation in mechanical energy:
−ΔP.E.=ΔK.E.
4πϵ01(1×106)2[10−31−9×10−31]=21×4×10−6×V2
9×109×10−12[9×10−38]=2×10−6×V2
V2=4×106
V=2×103ms−1
JEE Main 2019 — Physics Electromagnetism
In free space, a particle A of charge 1μC is held fixed at point P . Another particle B of the same charge and mass 4μg is kept at a distance of 1mm from P. If B is released, then its velocity at a distance of 9mm from P is:
[Take 4πϵ01=9×109Nm2C−2 ]
Held on 10 Apr 2019 · Verified 6 Jul 2026.
1.0ms−1
1.5×102ms−1
2.0×103ms−1
3.0×104ms−1
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