Verified Solution • 2025–26 Session moderate

QJEEM25Ph7aeb: JEE Main 2025 Physics - Class 12 JEEM25Ph7aeb Solution: A small uncharged conducting sphere is placed in contact with…

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−8C4 \times 10^{-8} \mathrm{C} charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N9 \times 10^{-3} \mathrm{~N} . The distance between them is (Take 14πϵo\frac{1}{4 \pi \epsilon_{\mathrm{o}}} as 9×1099 \times 10^9 in SI units)

Multiple Choice Options

  • A 1 cm
  • B 2 cm
  • C 4 cm
  • D 3 cm

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Evening Shift Physics - Electrostatics Question 4 English Explanation

F=k(θ2)(θ2)r29×10−3=9×109×(4×10−8)×4×10−84×r2r2=9×109×16×10−164×9×10−3=4×10−4r=2×10−2 m⇒2 cm

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