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
- A7 μ C , 3 μ C
- B8 μ C , 2 μ C
- C5 μ C , 5 μ C
- 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