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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)

Q61JEE Main 2021 · 25 FEB · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is a21×10−8\frac{a}{21} \times 10^{-8} C. The value of 'a' will be ___________. [Given g = 10 ms−2{}^{-2}]

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

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Answer: 20

  1. Half separation 0.1 m0.1\text{ m}, thread 0.5 m0.5\text{ m}: tan⁡θ=0.10.25−0.01=126≈0.204\tan\theta = \dfrac{0.1}{\sqrt{0.25 - 0.01}} = \dfrac{1}{2\sqrt{6}} \approx 0.204.
  2. F=mgtan⁡θ=10−5×10×0.204=2.04×10−5 NF = mg\tan\theta = 10^{-5} \times 10 \times 0.204 = 2.04 \times 10^{-5}\text{ N}
  3. q2=Fr2k=2.04×10−5×0.049×109=9.07×10−17⇒q=9.52×10−9 Cq^2 = \dfrac{Fr^2}{k} = \dfrac{2.04 \times 10^{-5} \times 0.04}{9 \times 10^9} = 9.07 \times 10^{-17} \Rightarrow q = 9.52 \times 10^{-9}\text{ C}
  4. a21×10−8=0.952×10−8⇒a=20\dfrac{a}{21} \times 10^{-8} = 0.952 \times 10^{-8} \Rightarrow a = 20.
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