What Is the Maximum Electrical Force Between Two Protons in a Cyclotron?

[stylez03]

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]

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.

 

[stylez03]

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]

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).

 

[stylez03]

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]

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

 

[stylez03]

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]

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.

 

[stylez03]

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 let's 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]

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

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.

 

[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

 

[Doc Al]

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

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:

$$A*D = B*C$$

 

[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?

 

[Doc Al]

Yes and yes.

 

[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?

 

[Doc Al]

... I'm wondering if km/s needs to be converted into m/s?

Absolutely.