Conservation of energy-Moving Charges

In summary, the problem involves 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 being projected head-on towards a gold nucleus with a charge of +79e. Using the conservation of energy equation and the equation for potential energy, the distance between the two nuclei can be calculated to be 4.83x10^-14 m.
  • #1
Moongurl767
15
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 :)
 
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  • #2
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
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
Yes, and what else?
 
  • #7
Ui is also zero
 
  • #8
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?
 
  • #9
(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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Related to Conservation of energy-Moving Charges

1. What is the principle of conservation of energy?

The principle of conservation of energy states that energy cannot be created or destroyed, but it can be transferred from one form to another. This means that the total amount of energy in a closed system remains constant.

2. How does the conservation of energy apply to moving charges?

In the context of moving charges, the conservation of energy means that the total energy of a charged particle remains constant as it moves through an electric or magnetic field. This energy can be in the form of kinetic energy, potential energy, or electromagnetic energy.

3. Can the energy of a moving charge change?

Yes, the energy of a moving charge can change as it interacts with electric or magnetic fields. For example, if a charged particle moves through an electric field, it will experience a change in potential energy. Similarly, if a charged particle moves through a magnetic field, it will experience a change in kinetic energy.

4. What is the relationship between the conservation of energy and the conservation of momentum?

The conservation of energy and the conservation of momentum are closely related. Both principles state that the total amount of a physical quantity (energy or momentum) in a closed system remains constant. In the context of moving charges, this means that the total energy and momentum of the charged particles in a system remain constant as they interact with electric and magnetic fields.

5. How does the conservation of energy affect the behavior of moving charges in an electric circuit?

In an electric circuit, the conservation of energy ensures that the total energy of the charged particles (electrons) remains constant as they move through the circuit. This means that the energy supplied by the power source is equal to the energy consumed by the various components in the circuit, such as resistors, capacitors, and inductors. As a result, the conservation of energy allows us to predict the behavior of the charges and the flow of electricity in a circuit.

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