Electrostatic potential of a split sphere, using the laplacian for spherical coord

[Vincent_111]

Homework Statement

Two concentric spherical shells. The outer shell is split into two hemispheres at potentials +Vo for the upper half and -Vo for the lower half. The inner shell is at zero potential (see attachment).

" what is the potential in the region; r > R' " (the potential in the space between the spheres is worked out as an example.)

Homework Equations

Pl(u) = legendre polynomial = P(u)l = 1/2^(l)l! * d^l/du^l (u^2 -1)^l


Integral,{bounderies; -1/1] Pl(u)Pn[/I](u) du = 0, if l is not n
Integral,{bounderies; -1/1] Pl(u)Pn[/I](u) du = 2/2l + 1, if l is n

general solutions of the potential in spherical coordinates:
F(r, angle a) = r^(l) Pl (cos a), r^(-l-1)Pl (cos a), l = 0,1,2,3,...

F(r, angle a) = the electrostatic potential dependent on radius(r) and one angle since axial symmetry is assumed

The Attempt at a Solution

the following attempt is a bit elaborate but in the end it comes down to finding the specific coefficient Am which has to be obtained form this formula:
Pm(cos a) sin a F(r', angle a) =
summ from l is 0 to infinity of [ {Al r'^(l) + Al r'^(-l-1)} Pl Pm(cos a) sin a(cos a) ]
If you are familiar with these kinds of problems/equations I would very much appreciate you input

There are two boundary conditions for the potential outside of the sphere:
1= the potential on the sphere which is given; +/-Vo.
2= the potential at a distance of infinity where the potential is zero.
We can plug these solutions at the bounderies in the general solution and try to find the specific potential that encompasses these two conditions.

F(r, angle a) = r^(l) Pl (cos a), r^(-l-1)Pl (cos a)
0 = summ from l= 0 to infinity of [ Al r^(l) Pl (cos a) + Bl r^(-l-1) Pl(cos a)]

I thought it might be possible ( as a physical 'trick' ) to get rid of the infinite radius since one is minus infinity and the other is plus infinity) then what we arer left with is:

0 = summ from l is 0 to infinity of [ Al Pl (cos a) + Bl Pl(cos a)]
since Pl (cos a) is not zero:

Al + Bl = zero

Bl = Al

Now we can put this value in the general soolution again and solve for the second boundary condition:

F(r', angle a) = summ from l is 0 to infinity of [ {Al r'^(l) + Al r'^(-l-1)} Pl (cos a) ]

we can multiply both sides with: Pm(cos a) sin a and then integrate from 0 to Pi (for angle a)


Pm(cos a) sin a F(r', angle a) =
summ from l is 0 to infinity of [ {Al r'^(l) + Al r'^(-l-1)} Pl Pm(cos a) sin a(cos a) ]

where we know that if we replace cos a by u that the second part of the equation on the right hand side is equal either to 0 or 2/2l+1 (see relevant formulas).

Now is where I am not sure how to continue (if it made sense up to this point);
we can find Al by shuffeling around the variables but how?
- the Pl(cos a) are replaced by 2/2l +1
- how does the summation over index l disappear?

 

[gabbagabbahey]

F(r, angle a) = r^(l) Pl (cos a), r^(-l-1)Pl (cos a)
0 = summ from l= 0 to infinity of [ Al r^(l) Pl (cos a) + Bl r^(-l-1) Pl(cos a)]

I thought it might be possible ( as a physical 'trick' ) to get rid of the infinite radius since one is minus infinity and the other is plus infinity) then what we are left with is:

Huh?

Are you claiming that \left.r^{-\ell-1}\right|_{r=\infty}=-\infty? Because it doesn't; one divided by an infinitely large number gives you zero, not negative infinity. Even if it were -\infty, you couldn't just pretend that \infty+-\infty=A_\ell P_\ell (\cos\theta)+B_\ell P_\ell (\cos\theta)... Adding two infinite numbers produces an undefined result.

You have that the potential is of the general form

\Phi(r,\theta)=\sum_{\ell=0}^{\infty}\left[A_\ell r^\ell +\frac{B_\ell}{r^{\ell+1}}\right]P_\ell(\cos\theta)

As r approaches infinity, r^\ell blows up (except for \ell=0). So the only way you can have a finite potential at r\to\infty is if all the A_{\ell}=0 except for possibly A_0. You should see, however, that \lim_{r\to \infty}\Phi(r,\theta)=A_0[/itex] and so A_0 also equals zero if you want the potential to be zero at infinity.