# Electrical Potential Force

## Homework Statement

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.

## Homework Equations

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

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

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

## The Attempt at a Solution

I'm not sure where to go from here.

Doc Al
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.

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.

What is V_i ?

Doc Al
Mentor
What is V_i ?
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).

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).

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

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

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

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

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

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

Doc Al
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.

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.

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

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

??

Doc Al
Mentor
$$KE_{i}1 + KE_{i}2 = PE_{f}$$
Good.

$$2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R}$$
Typos in the right hand term.

$$2(m_{p}*v_{i}) = \frac {KQ^{2}} {R}$$
Almost--that should be v_i^2. And the 2 cancels.

We divide out KQ^2 to isolate R

$$\frac {2(m_{p}*v_{i})} {KQ^{2}} = {R}$$
Do this step over. Note that R starts out in the denominator.

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

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

Doc Al
Mentor
$$m_{p}*v_{i} = \frac {KQ^{2}} {R}$$
Good.

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

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

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

Multiply by R

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

Sorry my algebra is a little rusty
No problem. Remember the cross-multiply trick. When you see this:

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

You can convert to this:

$$A*D = B*C$$

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?

Doc Al
Mentor
Yes and yes.

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

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

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

$$R = 0.000051$$

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

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

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

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:
Doc Al
Mentor
... I'm wondering if km/s needs to be converted into m/s?
Absolutely.