# Electric Potential (Potential Energy)

1. Feb 11, 2009

### erinec

1. The problem statement, all variables and given/known data
A proton is fired from far away towards a Hg nuclide.
Determine the distance of closest approach of the proton to the
centre of the nuclide when the initial speed of the proton is
4.0 * 107 m/s.

2. Relevant equations
(1/2)mv2 = kqq/r

3. The attempt at a solution
I plug the numbers straight in and solve for r to get 3.113 * 10^-15, which is totally different from the correct answer above. (For mercury charge, I used 80e since mercury has 80 protons.)

2. Feb 11, 2009

### AEM

Check each term in your expression. Are you using consistent units? That seems like a likely place for error.

3. Feb 11, 2009

### erinec

Yes I used consistent units. And still not getting the answer.
I do not know what I am missing.

4. Feb 11, 2009

### AEM

Well, I just computed it out using the following values:

Mass of proton = 1.67 X 10 ^-27 kg

electronic charge = 1.6 X 10^-19 coul

$$K = \frac{1}{4\pi\epsilon_o}$$ with $$\epsilon_o$$ = 8.85 X 10^-12 farad/meter and the rest as you gave and got the right answer. So put it aside and try it in the morning. It could be a simple calculation error.

5. Feb 11, 2009

### erinec

May I ask what formula you are using to solve for the distance?

6. Feb 11, 2009

### AEM

Well, I just computed it out using the following values:

Mass of proton = 1.67 X 10 ^-27 kg

electronic charge = 1.6 X 10^-19 coul

$$K = \frac{1}{4\pi\epsilon_o}$$ with $$\epsilon_o$$ = 8.85 X 10^-12 farad/meter and the rest as you gave and got the right answer. So put it aside and try it in the morning. It could be a simple calculation error.

7. Feb 11, 2009

### AEM

Sorry for the double post. I used your equation that you posted. After all, it is the correct expression. Since energy is conserved all of the kinetic energy of the particle become potential energy of the particle at its point of closest approach. I solved your equation for r just like you said.

8. Feb 11, 2009

### erinec

Ah thanks I got the correct answer now.

9. Feb 11, 2009

### AEM

You're welcome!