How Does Electron Distribution Affect Electric Fields in a Uranium Atom?

[xxbigelxx]

Homework Statement

The electrons of the uranium atom can be approximated as a uniform density sphere with radius 7.4 x 10-13 cm. The nucleus is a point at the center with a charge of +92e, and there are 92 electrons.
a. Calculate the electric potential and electric field at the surface taking only the electrons into account.
b. Calculate the electric potential and electric field at the surface taking only the nucleus into account.
c. Compare the 2 sets of values. d. A more classical picture has the electrons in planetary orbits. At any instant, there are 92 electrons arrayed uniformly about the sphere. Discuss how this changes the answers to the 3 previous parts.

Homework Equations

The Attempt at a Solution

I am not entirely understanding the setup for this problem. I am having a hard time understanding what is being described. If I am understanding it correctly, I would guess I would have to use Gauss Law to get the value for E, and then use that value to find V. Can anyone comment on my thought process?

 

[diazona]

That's one way to go about it. Try it and see if you can get it to work.

 

[xxbigelxx]

Well wouldn't the values for E be:
due to electrons only...(-92e/epsilon-naught)(4Pi*r^2) where r= 7.4 x 10-13 cm?
due to nucleus only... the same as above, except it would be positive?

 

[xxbigelxx]

I also don't know how to find the potential (and I am not sure if my above post was correct either).

 

[diazona]

OK, well your method for the electric field sounds right. What do you know that might allow you to calculate the potential?

 

[xxbigelxx]

I was thinking about gradient of E is equal to V?

 

[diazona]

It's the other way around,
$$\vec{\nabla}V = \vec{E}$$

 

[xxbigelxx]

Oh right that's what I meant (I was typing a bit too quickly). I thought about this technique but I wasn't sure how it would work. I tried a few things on scratch paper, but I kept hitting dead ends.

 

[diazona]

The equation as given allows you to calculate E from V by taking the derivative, but you need to go the other way around: you need to calculate V from E. How would you rearrange the equation to do that?

 

[xxbigelxx]

Would you have to integrate?

 

[diazona]

Yep. Can you write out the formula in full?
$$V(\vec{r}) = \int_{?}^{?}\cdots$$

 

[xxbigelxx]

Isn't it just the integral of E.dl from where you are measuring it to where the reference point is?

 

[diazona]

Yep, that's correct. How would you apply that here?

 

[xxbigelxx]

That's where I'm getting stuck. Isn't the value of E going to be a constant at the edge of the atom? Would it just be that constant value of E times the Radius (dl)?

 

[diazona]

Let's back up a bit: where is your reference point? And where are you measuring the field? I think you need to first be sure you know where the two endpoints of the integral are.

(By the way, the integral goes from the reference point to the location where you're measuring the potential, not the other way around.)

 

[xxbigelxx]

Would the reference point be at infinity, since the potential there is 0? And we are finding the field at the given value of 'r.' I hope this is right.

 

[diazona]

That's right. So you need to integrate the electric field from infinity to the electron sphere's radius, which I'll denote r_e for simplicity (rather than writing out the number every time).
$$\int_\infty^{r_e} \vec{E}(\vec{r})\cdot\mathrm{d}\vec{l}$$
Can you do something with that?

 

[xxbigelxx]

Well I believe that E is constant, so that can be pulled to the front of the integral. That leaves you with E(l1-l2) where l1=r=7.4 x 10-13 cm. What about that infinity? That seems to change things...

 

[diazona]

Well, that's incorrect. E is not constant. Why did you think it was?

Can you write a formula for E as a function of position?

 

[xxbigelxx]

I thought it was since earlier I said that E was equal to (-92e/epsilon-naught)(4Pi*r^2) where r= 7.4 x 10-13 cm. I thought all of those values were constant at the surface of the sphere, and E was constant as a result.

 

[diazona]

Ah, well you found the value of E at the surface of the sphere - that is, you found it at one particular location. That doesn't mean it's constant. In order for E to be constant, it would have to have that value at all locations in space.

Do you understand that in order to do the integral, you need to have values for E not only at the surface of the sphere, but all the way out to infinity?

 

[xxbigelxx]

Ohh I think I understand now. So my integral would be Edl where E=-92e/epsilon-naught)(4Pi*r^2) and dl=dr? Also the bounds for r are from infinity to the given value of r?

 

[diazona]

Yeah, that's more like it.

 

[xxbigelxx]

Ok so I got V=-92e/[(1.5E-35)*epsilon-naught]

 

[diazona]

OK, well if you did the math correctly, I suppose that would be the answer. (I haven't calculated it myself)

Although: make sure you include the correct units, and think about whether the sign of the result (positive or negative) makes sense.

 

[xxbigelxx]

My instructor just posted the solutions for these problems, and this is what he did for the value of E. He didn't even compute an integral. He just said V=k*-92e/7.4E-15 and E=k*-92e/(7.4E-15)^2. I don't understand how he used these values of the distance. Doesn't that value change for the electrons?