# Electrical Potential Force

1. Feb 24, 2007

### stylez03

1. The problem statement, all variables and given/known data
Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1350 km/s, measured relative to the earth.

Find the maximum electrical force that these protons will exert on each other.

2. Relevant equations

Mass of Proton $$1.67 * 10 ^{-27}$$

$$KE_{i} = PE_{f}$$

$$\frac {1} {2} * m_{p}*v_{1}^{2} = \frac {1} {4*pi*8.85*10^{-12}}*\frac {Q_{2}} {R}$$

3. The attempt at a solution

I'm not sure where to go from here.

2. Feb 25, 2007

### Staff: Mentor

Find the distance of closest approach by setting the initial KE (total KE) equal to electric potential energy. Then use Coulomb's law to find the force.

3. Feb 25, 2007

### stylez03

What is V_i ?

4. Feb 25, 2007

### Staff: Mentor

What do you mean by V_i? The initial speed? That's given. The initial potential? Assume they start out infinitely far away from each other (for all practical purposes).

5. Feb 25, 2007

### stylez03

To find the closet approach, I'm solving for R as you said so it should be in the form of:

$$\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}} = R$$

6. Feb 25, 2007

### Staff: Mentor

Looks to me like you are only counting the KE of one proton--use the total KE of both.

7. Feb 25, 2007

### stylez03

When we count both, would it just be KE_1 + KE_2?

$$2*(\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}}) = R$$

8. Feb 25, 2007

### Staff: Mentor

That's right--just add up the KE. But fix that equation--I just realized that you have some factors inverted. The magnitude of the PE is kq^2/R.

9. Feb 25, 2007

### stylez03

Ah lets show each step:

$$KE_{i}1 + KE_{i}2 = PE_{f}$$

$$2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R}$$

$$2(m_{p}*v_{i}) = \frac {KQ^{2}} {R}$$

We divide out KQ^2 to isolate R

$$\frac {2(m_{p}*v_{i})} {KQ^{2}} = {R}$$

??

10. Feb 25, 2007

### Staff: Mentor

Good.

Typos in the right hand term.

Almost--that should be v_i^2. And the 2 cancels.

Do this step over. Note that R starts out in the denominator.

Last edited: Feb 25, 2007
11. Feb 25, 2007

### stylez03

$$m_{p}*v_{i} = \frac {KQ^{2}} {R}$$

Divide by $$m_{p}*v_{i}$$

$$\frac {KQ^{2}} {(m_{p}*v_{i})*R}$$

Multiply by R

$$R = \frac {KQ^{2}} {(m_{p}*v_{i})}$$

Sorry my algebra is a little rusty

12. Feb 25, 2007

### Staff: Mentor

Good.

You mean:

$$1 = \frac {KQ^{2}} {(m_{p}*v_{i})*R}$$

Good.

No problem. Remember the cross-multiply trick. When you see this:

$$\frac{A}{B} = \frac{C}{D}$$

You can convert to this:

$$A*D = B*C$$

13. Feb 25, 2007

### stylez03

After I find the closet approach, you said to use Coulomb's Law to find the force. Is the R in here equilv to d in Coulomb's Law where:

$$F = \frac {K*Q_{1}*Q_{2}} {d^{2}}$$

and F is what I'm after in t his problem?

14. Feb 25, 2007

### Staff: Mentor

Yes and yes.

15. Feb 25, 2007

### stylez03

$$R = \frac {KQ^{2}} {(m_{p}*v_{i}^{2})}$$

$$R = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(1.67*10^{-27})*(1350)^{2}}$$

$$R = \frac {2.30*10^{-28}} {4.50*10^{-24}}$$

$$R = 0.000051$$

$$F = \frac {K*Q_{1}*Q_{2}} {d^{2}}$$

$$F = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(0.000051)^{2}}$$

$$F = 8.84 * 10^{-20}$$

Can someone verify my calculations, the online program says it's incorrect. I think I did everything right, I'm wondering if km/s needs to be converted into m/s?

Last edited: Feb 25, 2007
16. Feb 26, 2007

Absolutely.