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JEE Main202313 Apr 2023Evening ShiftPhysicsElectrostaticsActual

A 10 μ C charge is divided into two parts and placed at 1 cm distance so that the repulsive force between them is maximum. The charges of the two parts are:

Options

  1. A7 μ C , 3 μ C
  2. B8 μ C , 2 μ C
  3. C5 μ C , 5 μ C
  4. D9 μ C , 1 μ C

Correct answer

C. 5 μ C , 5 μ C

Step-by-step solution

Let the charges be x   μC ,   q - x   μC , where q = 10   μC The force between them is F = K x ( q - x ) r 2 . For the force to be maximum, d F d x = 0 ⇒ d F d x = K ( q - 2 x ) r 2 = 0 ⇒ x = q 2 = 5   μC

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