Problem on Electric Field

1. Aug 31, 2005

echau

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.

2. Aug 31, 2005

Staff: 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.

3. Aug 31, 2005

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.

4. Aug 31, 2005

echau

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 =/

5. Aug 31, 2005

Staff: Mentor

The two approaches are identical. Take your pick.

6. Aug 31, 2005

echau

thank you :) i really appreciate the help!

7. Sep 1, 2005

lightgrav

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

8. Sep 1, 2005

Staff: 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.