Q64 / 90 · JEE Main 2023 (13 Apr Shift 2) · Physics

A 10 μC  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:

  • A
    7μC,3μC
  • B
    8μC,2μC
  • Correct answer
    C
    5μC,5μC
  • D
    9μC,1μC

Solution

Let the charges be x μC, q-x μC, where q=10 μC

The force between them is 

F=Kx(q-x)r2.

For the force to be maximum,

dFdx=0⇒dFdx=K(q-2x)r2=0⇒x=q2=5 μC

Hence, the correct answer is option: 'C'.

Relevant Questions