F=r2k(2θ)(2θ)9×10−3=4×r29×109×(4×10−8)×4×10−8r2=4×9×10−39×109×16×10−16=4×10−4r=2×10−2 m⇒2 cm
JEE Main 2025 — Physics Electromagnetism
A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−8C charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N. The distance between them is (Take 4πϵo1 as 9×109inSI units)
Held on 24 Jan 2025 · Verified 6 Jul 2026.
3 cm
2 cm
4 cm
1 cm
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