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Electric fields and potentials of a point charge surround by a charged spherical shel

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Homework Statement


A point charge +Q is surrounded by a spherical shell of inner radius a and outer radius b. The spherical shell has charge density [tex]\alpha[/tex]r.
a) What is the electric field for r < a?
b) What is the electric field for a < r < b?
c) What is the electric field for r > b?

d) What is the electric potential for r < a?
e) What is the electric potential for a < r < b?
f) What is the electric potential for r > b?



Homework Equations





The Attempt at a Solution



For practice, I'm trying to do the problem in two ways: finding the electric fields first and then integrating to get the potential, and finding the potential first and differentiating to get the fields. But it seems that I'm stuck with both ways:

I've already computed the electric fields (using Gauss' Law). Here they are:
r < a: Q / (4*Pi*Epsilon*r^2)
a < r < b: (Q + (1/2)*[tex]\alpha[/tex]*(r^2-a^2)) / (4*Pi*Epsilon*r^2)
r > b: Q + (1/2)*[tex]\alpha[/tex]*(b^2-a^2)) / (4*Pi*Epsilon*r^2)

To find the potentials, I thought I could simply integrate the fields from r to Infinity. This works in the first case (r < a), giving me a potential of Q / (4*Pi*Epsilon*r)
However, this is as far as it goes: the integral for the second case (a < r < b) doesn't converge if I use r and Infinity as bounds again. I assume I need to use different bounds, but I'm still a bit unclear on how exactly this works.

The second way, finding the potential first, is even harder. I assume it would be
k * Integral(dq/r), but I can't figure out how to express the differential element dq.

Thanks for any hints.
 

Answers and Replies

  • #2
ehild
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The potential is zero at infinity. So start to integrate the electric field strength from r>=b to infinity. From that you get the potential at b. Now integrate from a<=r<=b, and get the potential at a. Last, integrate from r<a to a. Is the charge density ρ=α*r? Then check your results for a<r<b and r>b.


ehild
 
  • #3
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The potential is zero at infinity. So start to integrate the electric field strength from r>=b to infinity. From that you get the potential at b. Now integrate from a<=r<=b, and get the potential at a. Last, integrate from r<a to a. Is the charge density ρ=α*r? Then check your results for a<r<b and r>b.
Hey ehild,

thanks for your answer. Here's what I still don't understand: I get that if I want to find the potential at b, I integrate from b to infinity. What I don't understand, however, is why I need to integrate from a to b to get the potential at a. Why not from a to infinity? After all, infinity is still the reference, right?

Also, do you have any pointers on how I could solve this problem the other way round, i.e. finding the potentials first?
 
  • #4
ehild
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You can integrate to infinity, but you need to integrate different functions in both domains. And you already know the integral from b to infinity, don't you? Why to calculate it again?

ehild
 
  • #5
SammyS
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What is the volume of the shell?

[tex]\frac{4\pi}{3}(b^3-a^3)[/tex]

What is the net charge on the shell?
 
  • #6
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@ehild: Oh, now it's making sense! Of course, the electric field function is a piecewise function in this case, I totally forgot that. Thanks a lot!
 
  • #7
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Well, turns out I'm still stuck. I've figured out the potential for a <= r < b, but it doesn't match the potential for r <= a at r=a. I double-checked all my integrals, so I don't think there's something wrong there. I'm starting to think that I messed up with the calculation for the electric fields.

Here's how I calculated the electric field for a < r < b, using Gauss' Law:

Q(enclosed) = Q + alpha * Integral(R, from a to r) = Q + (1/2)*alpha * (r^2 - a^2).

E*A = Q(enclosed)/epsilon

==> E = (Q + (1/2)*alpha * (r^2 - a^2)) / (4*Pi*epsilon*r^2)

Is this correct?
 
  • #8
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Ah, I found the mistake I made in calculating the electric fields. Q(enclosed) should be
Q + alpha * Integral(4 * r^3 * Pi, from a to r).

However, I still don't get the correct answer for the potential at a < r < b. Here's what I did:

Integrate((Q + alpha * Pi * (r^4 - a^4)) / (4 * Pi * eps * r^2)), from r to b) + Q / (4 * Pi * eps * r^2)

This gives me:

Q / (4 * Pi * eps * r) + (alpha * (b^3 - r^3)) / (12 * eps). If I stick in a for r, I should get Q / (4 * Pi * eps * a), which I don't. So it seems that my bounds are still wrong...

Any hints? I've tried this problem in so many ways now that my head is burning...

Thanks.
 
  • #9
ehild
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The potential inside the shell is U(r)=U(b)+integral(Edr) from r to b.
The potential is not Q / (4 * Pi * eps * r) for r<=a. Ir would be without the outer charges. You are free to chose the potential at a point, so you can assume that it is Q / (4 * Pi * eps * a) at r=a, but then it is not zero at infinity.

ehild
 
Last edited:
  • #10
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Thanks for your answer, ehild. I still don't see though why the potential isn't / (4 * Pi * eps * r) for r <= a. Here's what I did:

- I found the electric field for r <= a first: Q / (4 * Pi * eps * r^2). I'm pretty sure this is right - after all, it's just Gauss' Law.

- Integrated this from r to infinity, which gives Q / (4 * Pi * eps * r). I just don't see the mistake...

Thanks again for your help. I appreciate your patience.
 
  • #11
ehild
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The electric field is correct. To get the potential, you have to integrate the electric field from r to infinity. But the electric field strength is not Q / (4 * Pi * eps * r^2) for r>a.

ehild
 
  • #12
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Understood. But it should be Q / (4 * Pi * eps * r^2) for r=a, correct? And since a is the "boundary point" between the potential function for r < a and the potential function for r > a, both functions should give me Q / (4 * Pi * eps * r^2) for r=a. Is that right?
 
  • #13
ehild
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But it should be Q / (4 * Pi * eps * r^2) for r=a, correct? Is that right?
It is right for the electric field strength.

The electric field strength is the negative gradient of the potential. You get the change of potential between two points by integrating between them:

∫dU (from r to a)=-∫Edr (from r to a)--->
U(r)-U(a) = ∫ Q / (4 * Pi * eps * r^2)dr (from r to a),

that is U(r)=Q / (4 * Pi * eps) * (1/r-1/a) +U(a).

If you choose U=0 at infinity, the potential at r<a is

∫E1dr (from r to a) +∫E2dr (from a to b)+∫E3dr (from b to infinity).

The potential at r=a is

∫E2dr (from a to b)+∫E3dr (from b to infinity)

ehild
 
  • #14
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Hi ehild,

thanks a lot for this. I've found another way to do it: compute the indefinite integrals of the negative of the electric field (once for each field). This gives you three constants of integration. You can get rid of the one for r > b, because it's 0. Now you know that V(r < a) and V(a < r < b) need to be equal when r = a. You also know that V(a < r < b) must equal V(r > b) for r = b. That gives you two equations that you can solve for the two integration constants.
Why that is equivalent to your way I don't understand yet. I'll take another look at it.

I do have one more question, though: we also needed to sketch the graph of the electric field with respect to r. How do I know if the piecewise graph from a to b is concave up or down? It almost seems like I have to curve analysis to find that out...
 
  • #15
ehild
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Both methods-using definite integrals, or indefinite ones and matching the constant - will give the same result.

The E(r) graph can be either concave and convex between a and b, depending on alpha and a, b. You can find it out from the second derivative.

For r>a the graph looks like the electric field of a point charge equal to the total charge of the system. The electric field between a and b is less steep as the charge decreases inward. For r<a, the graph is the same as that of a point charge Q + the additive term so as it matches to E(a) at r=a.

ehild
 

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