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JEE Main 2022 (24 Jun, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2022 (24 Jun, Shift 2)

Q48JEE Main 2022 · 24 JUN · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two identical charged particles each having a mass 10 g and charge 2.0 ×\times 10−-7C are placed on a horizontal table with a separation of L between them such that they stay in limiting equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms−2{}^{-2}]

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Answer: (a)

  1. Limiting equilibrium: kq2L2=μmg\dfrac{kq^2}{L^2} = \mu mg.
  2. kq2=9×109×4×10−14=3.6×10−4kq^2 = 9 \times 10^9 \times 4 \times 10^{-14} = 3.6 \times 10^{-4}; μmg=0.25×0.01×10=0.025 N\mu mg = 0.25 \times 0.01 \times 10 = 0.025\text{ N}.
  3. L2=3.6×10−40.025=0.0144 m2⇒L=0.12 mL^2 = \dfrac{3.6 \times 10^{-4}}{0.025} = 0.0144\text{ m}^2 \Rightarrow L = 0.12\text{ m}
  4. So the answer is (a) 12 cm.
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