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← JEE Main 2022 · 24 Jun · Shift 2
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JEE Main 2022 · 24 Jun · Shift 2, Physics question 34

Two identical charged particles each having a mass 10 g and charge 2.0×10−72.0\times10^{-7} C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g=10 ms−2g=10\ \mathrm{ms^{-2}}]

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Answer and worked solution

Official answer: A

Limiting equilibrium. Coulomb repulsion equals the limiting friction: kq2L2=μmg\dfrac{kq^2}{L^2}=\mu mg.

Numbers. 9×109×(2×10−7)2L2=0.25×0.01×10=0.025\dfrac{9\times10^9\times(2\times10^{-7})^2}{L^2}=0.25\times0.01\times10=0.025, so L2=3.6×10−40.025=0.0144L^2=\dfrac{3.6\times10^{-4}}{0.025}=0.0144 m2^2.

So L=0.12L=0.12 m =12=12 cm.

(A)

Topics:Coulomb's law and superpositionFriction
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