Potential inside a hollow sphere

  • #1
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I have a problem finding the potential inside a hollow sphere. The problem is formulated as:
"Determine the electric potential inside a hollow sphere of radius a given that the potential on its surface is f(theta)"
I have found a function V(r,theta) through separation of variables and all I have left is to apply the boundary conditions. One is given, but I guess I need another one when my function has two unknown constants. How do I find that second boundary condition?
 

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  • #2
siddharth
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Your radial solution will be of the form

[tex] f(r) = Ar^n + B/r^{n+1} [/tex]

So, what happens when r goes to 0? What does that tell you about B?

You'll need only one boundary condition for [itex] V(r,\theta) [/itex] as a result. (Although you'll have infinite constants. Use the orthogonality property of Legendre polynomials to calculate them)
 
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  • #3
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So this forces B to equal zero, right? Then my potential function looks like V(r,theta) = A_n r^n P-n(cos(theta)) and I should somehow use that the integral over P_m* P_n equals zero if m is different from n, ok? I just don't know how..?
(Damn, it would be easier if I knew how to use LaTeX) ;)
 
  • #4
siddharth
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Logarythmic said:
So this forces B to equal zero, right?
Yeah

Then my potential function looks like V(r,theta) = A_n r^n P-n(cos(theta)) and I should somehow use that the integral over P_m* P_n equals zero if m is different from n, ok?
I just don't know how..?
(Damn, it would be easier if I knew how to use LaTeX) ;)

You've almost got it. Your potential function will be
[tex] V(r,\theta)= \sum_{n=0}^\infty A_n r^n P_n(\cos \theta) [/tex]

and so at r=R

[tex] f(\theta)= \sum_{n=0}^\infty A_n R^n P_n(\cos \theta) [/tex]

(Check out https://www.physicsforums.com/showthread.php?t=8997", or click the image to see the code)

Since the legendre polynomials are orthogonal, multiply both sides of the above equation by [tex] P_m (\cos \theta) \sin \theta [/tex] (this is important. Do you see why?) and integrate (can you tell me the limits?). So, can you take it from here?
 
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  • #5
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I understand why I should multiply both sides with P_m(cos theta) but not why to include sin theta. The limits should be from 0 to pi, right?
 
  • #6
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I got it now. sin(theta) is for the substitution cos(theta) = x to work out. Then I get an expression for A_m with an integral over P_m(cos theta)f(theta)sin(theta)d(theta). But then what?
 
  • #7
siddharth
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Logarythmic said:
I got it now. sin(theta) is for the substitution cos(theta) = x to work out. Then I get an expression for A_m with an integral over P_m(cos theta)f(theta)sin(theta)d(theta). But then what?

Yes. Now, what about the other side of the equation? Remember that your aim is to get the value of [itex]A_n[/itex].

If you found that as an integral, then you're done! You know the solution to the potential at any point in the sphere.
 
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  • #8
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But the correct answer should be
[tex]V(r,\theta) = -\frac{u_0}{3} + \frac{4u_0}{3}(\frac{r}{R})^2P_2(\cos \theta)[/tex].

The given potential at R is
[tex]V(R,\theta) = u_o \cos(2\theta)[/tex].

(I'm using LaTeX!)
 
  • #9
siddharth
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Since you know [itex]f(\theta)[/itex], evaluate the integral using the orthogonal property!

Where exactly are you stuck at? Can you post that here?
 
  • #10
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I substituted [tex]\cos \theta = x[/tex] and used the orthogonal property. Then I re-substituted to get the [tex]\theta[/tex]-dependence again and now my expression looks like

[tex]A_m = -\frac{1+2m}{2} a^\textit{-m} \int_{0}^{\pi} P_m(\cos \theta) u_0 cos(2\theta) sin \theta d\theta[/tex]

I dunno where to go from here.
 
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  • #11
siddharth
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To simplify the integral on the RHS, as I've said before, you're going to need the orthogonal property . Have you tried writing [itex]\cos(2\theta) [/itex] in terms of [itex]P_2(\cos \theta) [/itex]? (Also, [itex]1=P_0(\cos \theta)[/itex])
 
  • #12
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Hey, I solved it! Thanks alot! This has really helped me in my understanding of Legendre polynomials.
 

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