Conservation of energy-Moving Charges

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Homework Help Overview

The discussion revolves around a conservation of energy problem involving a helium nucleus and a gold nucleus, focusing on the interaction between their charges and the calculation of distance before the helium nucleus stops and reverses direction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the conservation of energy equation, questioning which initial and final quantities are zero. They discuss the kinetic and potential energy expressions and the relationship between them.

Discussion Status

Participants are actively engaging with the problem, clarifying concepts and working through calculations. Guidance has been provided on how to rearrange equations and interpret the energy conservation principle, leading to a clearer understanding of the steps involved.

Contextual Notes

There is an emphasis on ensuring the correct interpretation of kinetic and potential energy, as well as the need to accurately rearrange equations to solve for the desired variable. Participants express uncertainty about their calculations and the implications of their results.

Moongurl767
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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 :)
 
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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?
 
The final velocity is zero.
 
You have four quantities in the energy conservation equation, Ki, Ui, Kf and Uf. Which one(s) do you think are zero?
 
Kf right?
 
Yes, and what else?
 
Ui is also zero
 
Moongurl767 said:
Ui is also zero
Correct because initially the particles are very far from each other. Now can you put the energy conservation equation together?
 
(1.50x10^7m/s)+0=0+Uf

Uf=1.50x10^7
 
  • #10
Moongurl767 said:
(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
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
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
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
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
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
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
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
 
  • #26
Right. Now divide both sides of the equation by Ki and you will get an expression with only r on the left side and a whole bunch of known quantities on the right.

Do the calculation and get your answer.
 
  • #27
Oh I see, that's where I went wrong, instead of dividing it I subtracted it. Okay so then this is what I get..

r=kq1q2/Ki

r=(9x10^9)(3.2x10^-19C)(1.26x10^.17C)/ (7.52x10^-13 J)r= 4.83x10^-14 m
 
  • #28
That's what I got. You're done. :smile:
 
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  • #29
Thank you! I'm glad you walked me through the problem instead of just giving me the answer, now I actually understand what I was doing! Thanks again!
 
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