r=10mm,x=2,
∣Fq∣=r2kλ⋅q
∣F−q∣=r+x2kλ⋅q
∣Fnet∣=r(r+x)2kλq⋅x
4=10mm.12mm2×9×109×3×10−6×q×2mm
⇒q=4.44μC
JEE Main 2021 — Physics Electromagnetism
An electric dipole is placed on x−axis in proximity to a line charge of linear charge density 3.0×10−6Cm−1. Line charge is placed on z−axis and positive and negative charge of dipole is at a distance of 10mm and 12mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.
Held on 22 Jul 2021 · Verified 6 Jul 2026.
815.1nC
8.8μC
0.485mC
4.44μC
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