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JEE Main 2021 (25 Feb, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2021 (25 Feb, Shift 2)

Q58JEE Main 2021 · 25 FEB · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two identical conducting spheres with negligible volume have 2.1 nC and −-0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is __________ ×\times 10−-9 N.

[Given: 4πε0=19×1094\pi\varepsilon_0 = \frac{1}{9 \times 10^{9}} SI unit]

Numerical answer type. Work it out, then open the answer.

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Answer: 36

  1. After contact each sphere has 2.1+(−0.1)2=1.0 nC\dfrac{2.1 + (-0.1)}{2} = 1.0\text{ nC}.
  2. F=kq2r2=9×109×(10−9)2(0.5)2F = \dfrac{kq^2}{r^2} = \dfrac{9 \times 10^9 \times (10^{-9})^2}{(0.5)^2}
  3. F=36×10−9 NF = 36 \times 10^{-9}\text{ N}. So the answer is 36.
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