Question

. A charge q is held fixed at the origin, and a charge q2 on a small ion of mass m is initially at fi travelling at velocity Ui. What is the closest that the ball with charge q2 gets to the origin, and what is its speed at that point? Assume that m 2.5 x 1025kg, 1 3.2 x 10-16C, 42 angular momentum around the origin is constant, and the condition for closest approach is F--:0. 4.8 x 10-19C, Ti 2.07 12, and

Please cleary demonstrate what would happen in this situation, and what path the charge would take. The correct answer for the speed at closest approach is 6.01 m/s and the seperation si 0.33 m.

Answer 1

: 275m N ow Cn) ai

So when the q2 charge starts approaching the charge q1 it starts experiencing a retardation..which is constant.

now this is due to the fact that they experience a electrostatic force given by the formula...

f (1/4pie(epsilon0))ql q2/r2

where r = distance between the 2 charges..

now the position vector of q1 initially is given...

so distance from the origin is ...

root(absicca ordinate2)

=root(10^2 + 1^2).