Electrical Potential Force

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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 [tex] 1.67 * 10 ^{-27} [/tex]

[tex] KE_{i} = PE_{f} [/tex]

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


The Attempt at a Solution



I'm not sure where to go from here.
 

Answers and Replies

  • #2
Doc Al
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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
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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 ?
 
  • #4
Doc Al
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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).
 
  • #5
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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:

[tex] \frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}} = R [/tex]
 
  • #6
Doc Al
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Looks to me like you are only counting the KE of one proton--use the total KE of both.
 
  • #7
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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?

[tex] 2*(\frac { \frac {1} {2}* m_{p} * v_{i}^{2}} {K*Q^{2}}) = R [/tex]
 
  • #8
Doc Al
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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
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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:


[tex] KE_{i}1 + KE_{i}2 = PE_{f} [/tex]

[tex] 2*(\frac {1} {2} * m_{p}*v_{1}^{2}) = \frac {1} {K}*\frac {Q_{2}} {R} [/tex]

[tex] 2(m_{p}*v_{i}) = \frac {KQ^{2}} {R} [/tex]

We divide out KQ^2 to isolate R

[tex] \frac {2(m_{p}*v_{i})} {KQ^{2}} = {R} [/tex]

??
 
  • #10
Doc Al
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[tex] KE_{i}1 + KE_{i}2 = PE_{f} [/tex]
Good.

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

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

We divide out KQ^2 to isolate R

[tex] \frac {2(m_{p}*v_{i})} {KQ^{2}} = {R} [/tex]
Do this step over. Note that R starts out in the denominator.
 
Last edited:
  • #11
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[tex] m_{p}*v_{i} = \frac {KQ^{2}} {R} [/tex]

Divide by [tex] m_{p}*v_{i} [/tex]

[tex] \frac {KQ^{2}} {(m_{p}*v_{i})*R} [/tex]

Multiply by R

[tex] R = \frac {KQ^{2}} {(m_{p}*v_{i})} [/tex]

Sorry my algebra is a little rusty
 
  • #12
Doc Al
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[tex] m_{p}*v_{i} = \frac {KQ^{2}} {R} [/tex]
Good.

Divide by [tex] m_{p}*v_{i} [/tex]

[tex] \frac {KQ^{2}} {(m_{p}*v_{i})*R} [/tex]
You mean:

[tex]1 = \frac {KQ^{2}} {(m_{p}*v_{i})*R} [/tex]

Multiply by R

[tex] R = \frac {KQ^{2}} {(m_{p}*v_{i})} [/tex]
Good.

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

[tex]\frac{A}{B} = \frac{C}{D}[/tex]

You can convert to this:

[tex]A*D = B*C[/tex]
 
  • #13
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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:

[tex] F = \frac {K*Q_{1}*Q_{2}} {d^{2}} [/tex]

and F is what I'm after in t his problem?
 
  • #14
Doc Al
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Yes and yes.
 
  • #15
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[tex] R = \frac {KQ^{2}} {(m_{p}*v_{i}^{2})} [/tex]

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

[tex] R = \frac {2.30*10^{-28}} {4.50*10^{-24}} [/tex]

[tex] R = 0.000051 [/tex]

[tex] F = \frac {K*Q_{1}*Q_{2}} {d^{2}} [/tex]

[tex] F = \frac {(8.99*10^9)*(1.60*10^{-19})^{2}} {(0.000051)^{2}} [/tex]

[tex] F = 8.84 * 10^{-20} [/tex]

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:
  • #16
Doc Al
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... I'm wondering if km/s needs to be converted into m/s?
Absolutely.
 

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