JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Electrostatics

Two identical charged particles each having a mass 10 g and charge

2.0 \times 1 0^{- 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

^{- 2}

]

(A)12 cm✓ Correct
(B)10 cm
(C)8 cm
(D)5 cm

Answer: (A)

Solution

At limiting equilibrium, the electrostatic repulsion equals the maximum friction force:

\frac{k q ^{2}}{L ^{2}} = μ m g

. Solving for L:

L = q \frac{k}{μ m g} = 2.0 \times 1 0^{- 7} \frac{9 \times 1 0 ^{9}}{0.25 \times 0.01 \times 10} = 2.0 \times 1 0^{- 7} \times 6 \times 1 0^{7} = 0.12

m = 12 cm.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option

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