A 10   μ C   charge is divided into two parts and placed at 1   cm distance so that the…

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:
  1. 7μC,3μC
  2. 8μC,2μC
  3. 5μC,5μC
  4. 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

Asked in: JEE Main 2023 (13 Apr Shift 2)

Practice more Electrostatics questions on Aicharya