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Conservation of energy-Moving Charges

  • #1

Homework Statement



If a helium nucleus with a mass of 6.68x10^-27 kg, a charge of +2e, and an initial velocity of 1.50x10^7 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)


Homework Equations



Ki+Ui=Kf+Uf

E=DeltaV/d a=Fe/m=eE/m=edeltaV/md


Vf=square root [Vi^2+2ad]





The Attempt at a Solution



I know this is a conservation of energy for moving charges problem but I just can't seem to figure out how to calculate the distance with these equations and the given information. I feel like I am completely missing something on this one! Help please :)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
kuruman
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It is conservation of energy indeed. Which ones (if any) of the initial and final quantities in the energy conservation equation is (are) zero?
 
  • #3
The final velocity is zero.
 
  • #4
kuruman
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You have four quantities in the energy conservation equation, Ki, Ui, Kf and Uf. Which one(s) do you think are zero?
 
  • #5
Kf right?
 
  • #6
kuruman
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Yes, and what else?
 
  • #7
Ui is also zero
 
  • #8
kuruman
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Ui is also zero
Correct because initially the particles are very far from each other. Now can you put the energy conservation equation together?
 
  • #9
(1.50x10^7m/s)+0=0+Uf

Uf=1.50x10^7
 
  • #10
kuruman
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(1.50x10^7m/s)+0=0+Uf

Uf=1.50x10^7
Where and how did you get this number? Energy is expressed in Joules. What equation did you use and what numbers did you plug in? I have to know exactly what you did in order to help you.
 
  • #11
This is where I'm having trouble. The 1.50x10^7 m/s is the Vi. I'm not sure how to get the energy.
 
  • #12
kuruman
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What is the expression that gives the potential energy of two charges separated by some distance r apart? Look it up if you don't remember.
 
  • #13
UE = k(q1q2/r) k= 9 x 10^9 N m2/C2
 
  • #14
kuruman
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That's the one. As we said before, you need to say

Uf = Ki

You already have that Uf = kq1q2/r

What expression can you write down for Kf? Look up "kinetic energy" if you have to.
 
  • #15
Kf=Ki+(Ui-Uf)

Kf=1/2mv^2
 
  • #16
kuruman
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OK. Can you find a number for this kinetic energy in Joules? You know everything that goes in the equation.
 
  • #17
Kf=(1/2)(6.68x10^-27kg)(1.50x10^7m/s)^2=

=7.52x10^-13 J
 
  • #18
kuruman
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What do you think you should do next?
 
  • #19
Well I'm a little confused as to why the equation is equal to Kf and not Ki because isn't Kf=0?
 
  • #20
kuruman
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You are correct, 7.52x10-13 J is Ki not Kf. Based on all that is said above, what is the next step?
 
  • #21
So then you would plug it into the equation Uf=Ki --> k (q1q2)/r=Ki

(7.52x10^-13 J)=(9x10^9 Nm^2/C^2) ((3.2x10^-19 C)(1.26x10^-17 C)/r)

and then solve for r to get r= 7.52x10^.13 m

I'm not sure if my math is correct though?
 
  • #22
kuruman
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The procedure is correct, but the number is not. Is it a coincidence that the number you got for r is the number you have on the left side for the kinetic energy?

Try doing the calculation again. Solve for r first in terms of the other quantities, then calculate. Post exactly what you did so that I can find any errors you may have made.
 
  • #23
I calculated it again and I still get the same thing. I've always had a hard time rearranging the equations in order to solve for a different variable. This is what I did...

Ki=k(q1q2)/r --> r=kq1q2-Ki

r= (9x10^9 Nm^2/C^2)*(3.2x10^-19C)*(1.26x10^.17C)-(7.52x10^-13 J)
r= (3.63x10^-26)-(7.52x10^-13)
r= (-7.52x10^-13)
 
  • #24
kuruman
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I see. Start with

Ki=kq1q2/r

If you move the kinetic energy to the other side and change sign, you get

0 = kq1q2/r - Ki

That's not what you want. Instead, multiply both sides of the equation with r. What do you get? (There is a second step, but I will tell you what it is after you complete this one.)
 
  • #25
rki=kq1q2
 
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