Question

In using classical mechanics to determine the Ruthoford model, why is coulombs law applied where the electron charge in the numerator is squared? If Z= atomic numberand e= electron charge, then why is the (q1q2) part of his equation (Ze^2) instead of (Ze)?

Answer 1

does the incident electron only get close enough to One of the targets electrons for the repulsive coulombs force to be regarded?

Answer 2

You're forgetting that Z is just a number, and the charge on the nucleus is Ze (Z times e, the charge on a single proton), not just Z. Hence the numerator of Coulomb's Law, which is q1 q2, is equal to (Ze)(-e) = - Ze^2, where Ze is the charge on the nucleus and -e is the charge on the electron.