Two electrons are fired at each other

[tobywashere]

Two electrons are fired at each other...

Homework Statement

Two electrons are fired at 3.5 x 10⁶ m/s directly at each other. Calculate the smallest possible distance between the two electrons.

Homework Equations

Ee = k(q1)(q2) / r

Ek = (0.5)mv²

The Attempt at a Solution

I looked at the problem a different way. I though of one electron as stationary, while the other electron moved towards the first at 7.0 x 10⁶ m/s (twice the speed of the single electron).
I let r represent the minimum distance between the particles. The work required to bring an electron from a distance of infinity to distance r is
k(1.6 x 10^-19)(1.6 x 10^-19) / r
(Charge on one electron is 1.6 x 10^-19 C)
(k is 9 x 10⁹)
The electron's original kinetic energy is converted into electrical potential energy. At the minimum distance apart, the kinetic energy is zero, so all the kinetic energy has been converted into electrical potential energy.
Thus,
(0.5)(9.11 x 10^-31)(7.0 x 10⁶)² = k(1.6 x 10^-19)(1.6 x 10^-19) / r
(mass of electron is 9.11 x 10^-31 kg)
Which means r = 1.0 x 10^-11 m
However, this answer is wrong. The actual answer is 4.5 x 10^-6 m. I can't figure out what I did wrong.

 

[Delphi51]

Thinking of only one electron moving, it would push on the other and the pair of electrons would still be moving at the time of closest approach and still have kinetic energy. You have to consider both moving, and work being done on both. I used 2*½mv² = 2*ke²/r
r = 2ke²/(mv²)
but end up with a bit less than that actual answer.

 

[cepheid]

Thinking of only one electron moving, it would push on the other and the pair of electrons would still be moving at the time of closest approach and still have kinetic energy. You have to consider both moving, and work being done on both. I used 2*½mv² = 2*ke²/r
r = 2ke²/(mv²)
but end up with a bit less than that actual answer.

I'm confused, because you end up with the same formula as the OP, only with v = 1/2 of what he or she used. And the OP's arithmetic was correct, so your modification just increases the OP's answer by a factor of 4, leaving it still more than 5 orders of magnitude too small.

 

[tobywashere]

I tried your method and I got 4.1 x 10^-11 m.
That's not a little bit off from the actual answer, that's a lot off...
But thanks for the suggestion.

 

[gneill]

Look at the problem in reverse. Suppose two electrons are held at rest close to each other with distance r separating them. What is the Potential energy of the system?

Well, we can forget gravity given the comparative strength of the electric field, so it's

PE = k*q²/r

That represents the work that would have to be done to bring the two particles together from infinity.

(We are of course ASSUMING that the fired electrons will be starting from essentially infinite distance. The problem doesn't explicitly say so, but it should, otherwise, what prevents us from assuming that the electrons are being fired from a half an angstrom apart at that initial speed?).

Divide the energy in half and give one half to each electron in the form of kinetic energy. The velocity and initial separation are related by:

\frac{1}{2} m_e v^2 = \frac{1}{2} k \frac{q_e^2}{r}

so, solving for r,

r = \frac{k q_e^2}{m_e v^2} = 2.07 \times 10^{-11} m

as was found earlier. I think that the "correct answer" isn't.

 

[nabaa]

I believe the equation for an Electric fields goes as follows:

kq1q2/r^2

not kq1q2/r

 

[nabaa]

this means you can equate

1/2kq1q2/r^2 = 1/2mv^2

multiply your answer by two because there are two charges. you cannot multiply by two initially or it will mess up the algebra, multiply your answer.

 

[cepheid]

I believe the equation for an Electric fields goes as follows:

kq1q2/r^2

not kq1q2/r

The relevant quantity here is electric potential energy, not electric field strength. The former goes like 1/r just as we said.

To everybody else: regardless of a few factors of two here and there, I think we all agree that the answer should be on the order of a few tens of picometres and NOT a few micrometres, and that the OP's answer key is just wrong, right?

 

[gneill]

I believe the equation for an Electric fields goes as follows:

kq1q2/r^2

That's for the force between two charges at separation r.

not kq1q2/r

That's for the Potential Energy due to the electric field for two charges at separation r.

 

[Delphi51]

Gneill is right; the PE is ke²/r, not double that as I wrote.
I was thinking you would have to push EACH charge in from infinity but apparently you only have to push one and assume the other is already there. Anyway, Wikipedia clearly states it is ke²/r.
http://en.wikipedia.org/wiki/Electric_potential_energy

Cepheid has summarized the situation perfectly.

 

[SammyS]

...

I looked at the problem a different way. I thought of one electron as stationary, while the other electron moved towards the first at 7.0 x 10⁶ m/s (twice the speed of the single electron).
...

Which means r = 1.0 x 10^-11 m

However, this answer is wrong. The actual answer is 4.5 x 10^-6 m. I can't figure out what I did wrong.

You can approach the problem this way IF you use the idea of reduced mass. In this case that means treating the moving electron as if its mass were m_e/2 . That gives the electron half the initial KE you used, giving a result consistent with that of gneill.

...

To everybody else: regardless of a few factors of two here and there, I think we all agree that the answer should be on the order of a few tens of picometres and NOT a few micrometres, and that the OP's answer key is just wrong, right?

You're right, it's just wrong !

 

[tobywashere]

So the textbook was wrong.
The actual answer is r = 2.06 x 10^-11
I know what the textbook did wrong: they took the square root of the correct answer, which gives 4.5 x 10^-6