# Problem on Electric Field

An alpha particle approaches a gold atom head on, stops, and turns around at a distance of 10^-11m from the nucleus. What is the electric field due to the gold nucleus at this point? Ignore the effects of the gold atom's orbiting electrons. What is the acceleration of the alpha particle when it is stopped? An alpha particle is a helium nucleus, composed of two protons and two neutrons.

Can anyone help me with this problem? I'm just not understanding it...any help would be appreciated.

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
The force momentarily stopping the alpha particle is the repulsive electric force that the gold nucleus exerts on the alpha particle. Use Coulomb's law to find the force between the charges at the given distance. (What's the charge of the gold nucleus? What's the charge of the alpha particle?) Then apply Newton's 2nd law to find the acceleration.

Astronuc
Staff Emeritus
The alpha particle (nucleus of He atom) has + charge proportional to Z=2 (2 protons) and the gold nucleus has + charge proportional to Z=79 (79 protons).

So this becomes an electrostatic force problem - the alpha stops.

Remember coulombs law and coulomb force.

What is the electric field cause due to 79q, where q is the magnitude of charge on a proton?

acceleration, a = F/m.

Thanks for the replies!

Should I use F=kq1q2/r^2? then plug it into F=ma?

or should I find E=kq/r^2 and plug it into F=qE?

Sorry if these are very simplistic questions...Physics is hard for me =/

Doc Al
Mentor
echau said:
Should I use F=kq1q2/r^2? then plug it into F=ma?

or should I find E=kq/r^2 and plug it into F=qE?
The two approaches are identical. Take your pick.

thank you :) i really appreciate the help!

lightgrav
Homework Helper
The two approaches are (pedagogically) NOT identical ...
the first approach ignores the E-field, which WAS the Question.

Doc Al
Mentor
Good point, since one of the questions was to find the electric field.

As far as figuring out the acceleration, the two methods for finding the force are identical. But since you have to find the electric field anyway, obviously you would use that result to finish the problem.