JEE Main 10 Apr 2019 Paper 2 Solved Paper

In free space, a particle A of charge 1µC is held fixed at a point P. Another particle B of the same charge and mass 4µg is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is:{Take

1

4πε0

=9×109Nm2C−2}