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JEE Main 2025 (24 Jan, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2025 (24 Jan, Shift 2)

Q22JEE Main 2025 · 24 JAN · SHIFT 2Medium

Electric Charge & Coulomb's Law

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−8 C4 \times 10^{-8}\text{ C} charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N9 \times 10^{-3}\text{ N}. The distance between them is (Take 14πϵ0\frac{1}{4\pi\epsilon_0} as 9×1099 \times 10^{9} in SI units)

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Answer: (b)

  1. After contact each sphere has q=2×10−8 Cq = 2 \times 10^{-8}\text{ C}.
  2. F=kq2r2⇒r2=9×109×4×10−169×10−3=4×10−4 m2F = \dfrac{kq^2}{r^2} \Rightarrow r^2 = \dfrac{9 \times 10^9 \times 4 \times 10^{-16}}{9 \times 10^{-3}} = 4 \times 10^{-4}\text{ m}^2
  3. r=2×10−2 mr = 2 \times 10^{-2}\text{ m}. So the answer is (b) 2 cm.
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