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JEE Main202224 Jun 2022Evening ShiftPhysicsElectrostaticsActual

Two identical charged particles each having a mass 10 g and charge 2 . 0 × 10 - 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 - 2 ]

Options

  1. A12   cm
  2. B10   cm
  3. C8   cm
  4. D5   cm

Correct answer

A. 12   cm

Step-by-step solution

Normal reaction applied by table on object will be, N = m g . Both charges are same, hence a repulsive force will act between them. Value of force acting will be, k Q 2 L 2 . Now the maximum value of friction force which table can provide is, = μ N = μ m g For equilibrium, k Q 2 L 2 = μ m g . L = k Q 2 μ m g = k 0 . 25 m g Q = 2 k m g Q = 2 9 × 10 9 10 × 10 - 3 × 10 2 × 10 - 7 = 0 . 12   m = 12   cm

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