Two electrons repel each other and gain velocity

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  • #1
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Homework Statement


Two electrons are held at rest 0.05[cm] apart. what is their relative velocity when they are 1[cm] apart.

Homework Equations


The potential V from a point charge: ##V=K\frac{q}{r}##
The constant ##K=9\cdot 10^9##
The work done to move from one point in the field to another: ##W=V\cdot q##
The electron charge: ##e=1.6\cdot 10^{-19}[Coulomb]##
The electron mass: ##m_e=9.11\cdot 10^{-31}[Kg]##

The Attempt at a Solution


The axes are attached to one of the electrons. the potential difference between the end and the beginning:
$$\Delta V=9\cdot 10^9\cdot 1.6\cdot 10^{-19}\cdot \left(\frac{1}{0.0005}-\frac{1}{0.01}\right)=2.7\cdot 10^{-6}$$
The work is the gained kinetic energy:
$$W=E_k=2.7\cdot 10^{-6}\cdot 1.6\cdot 10^{-19}=4.378\cdot 10^{-25}=\frac{1}{2}\cdot 9.11\cdot 10^{-31}\cdot V^2\rightarrow V=0.0023[m/s]$$
It should be 982[m/s]
 

Answers and Replies

  • #2
Doc Al
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The work is the gained kinetic energy:
$$W=E_k=2.7\cdot 10^{-6}\cdot 1.6\cdot 10^{-19}=4.378\cdot 10^{-25}=\frac{1}{2}\cdot 9.11\cdot 10^{-31}\cdot V^2\rightarrow V=0.0023[m/s]$$
Redo that last bit of arithmetic. Also realize that the KE is shared by both electrons.
 
  • #3
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It comes out V=980, it's good. i sit on one of the electrons and watch the other, the frame is attached to one electron so only the second moves, so KE is only of one electron. i know both move at the same speed but from the point of view of one of them only the other moves, so i guess V=980 is the answer and my way is good, no?
 
  • #4
Doc Al
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i sit on one of the electrons and watch the other, the frame is attached to one electron so only the second moves, so KE is only of one electron.
Realize that if you try to use the final frame of one of the electrons that the initial speeds are not zero. Don't do that. Stick to the original inertial frame in which the electrons were at rest.

i know both move at the same speed but from the point of view of one of them only the other moves, so i guess V=980 is the answer and my way is good, no?
Again I suggest finding the speed of each electron in the original frame by finding the KE of each electron. Then compute the relative speed.
 
  • #5
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Realize that if you try to use the final frame of one of the electrons that the initial speeds are not zero. Don't do that. Stick to the original inertial frame in which the electrons were at rest.
The electron is in rest also in the non inertial frame since the electrons are at rest in the beginning.
I suggest finding the speed of each electron in the original frame by finding the KE of each electron.
I don't know how to do that. i think i made it in my calculation, which gave V=980. if so then each one moves at that speed and the relative speed is double, but the book says only 982.
 
  • #6
Doc Al
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The electron is in rest also in the non inertial frame since the electrons are at rest in the beginning.
You cannot use an accelerating frame. Stick to one inertial frame. (Which is what you did anyway, despite what you think you did.)

I don't know how to do that. i think i made it in my calculation, which gave V=980.
You found the total KE (in the original frame) and attributed that total energy to one of the electrons. Instead, split it up between the two.

if so then each one moves at that speed and the relative speed is double, but the book says only 982.
When you find the correct electron speed, you'll get the correct relative velocity.
 
  • #7
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Because of symmetry:
$$W=E_k=4.378\cdot 10^{-25}=2\cdot \left(\frac{1}{2}\cdot 9.11\cdot 10^{-31}\cdot V^2\right)\rightarrow V=693[m/s]$$
Too high
 
  • #8
Doc Al
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Because of symmetry:
$$W=E_k=4.378\cdot 10^{-25}=2\cdot \left(\frac{1}{2}\cdot 9.11\cdot 10^{-31}\cdot V^2\right)\rightarrow V=693[m/s]$$
Too high
I agree with that calculation. So, unless we both made the same arithmetic error, your book is wrong.

Is this a textbook problem? Or a problem created by your instructor?
 
  • #9
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An old text book of 30 years, it's Sears Zemansky translated. i have much appreciation to it, the pages are brown.
If the charges weren't equal in mass and charge, i guess energy itself wouldn't suffice, right?
 
  • #10
Doc Al
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An old text book of 30 years, it's Sears Zemansky translated. i have much appreciation to it, the pages are brown.
Ah yes, a classic text. The current version taken over by Young and Freedman.

If the charges weren't equal in mass and charge, i guess energy itself wouldn't suffice, right?
If the masses were not equal, you'd have to invoke conservation of momentum to see how the energy divides between the two particles. Here conservation of momentum is trivial.
 
  • #11
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Thanks
 

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