What Is the Method for Finding the Potential of a Spherical Dipole?

In summary, the homework statement is that a spherical shell (radius R) carrying uniform charge sigma0 on the "northern" hemisphere, –sigma0 on the "southern" one. Find the potential inside and outside the sphere, calculating coefficients explicitly up to A6 and B6.
  • #1
Goddar
205
16

Homework Statement


(Griffith' electrodynamics, problem 3.22): a spherical shell (radius R) carries uniform charge sigma0 on the "northern" hemisphere, –sigma0 on the "southern" one. Find the potential inside and outside the sphere, calculating coefficients explicitly up to A6 and B6.


Homework Equations


Attached document: i don't know how to paste equations on this page, it's my first time.. for reference, it's the same basic solution developed for example 3.9 of the book, p.142


The Attempt at a Solution


the above makes reference the method of separation of variables (in spherical coordinates), with solution in the form of a Fourier series with coefficients involving legendre polynomials (variable = "cosine theta"): since the charge density is constant over each hemisphere, i end-up with no polynomial at all.. While an infinite series seems expected (this would be more clear if i could paste my equations); i guess i don't know where to plug-in the right boundary condition.. Anybody could enlight me?
 

Attachments

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  • #2
For a tutorial in LaTeX coding:

https://www.physicsforums.com/showthread.php?t=8997

Well, in getting to the point where you are at, you have already used all the boundary conditions that you can (maybe you should review what Griffiths does to get to the A_l and B_l correlation).

The good news is that you're doing fine. Since it is crappy to integrate a legendre polynomial between 0 and pi, you will want to use a trick to change it so it is between -1 and 1. If you want to see it then look down, but if you want to work on it yourself then don't look.


[tex]A_l = \frac{\sigma_0}{2 \epsilon_0 R^{l-1}}(\int_0^{\pi/2}P_l(cos \theta)sin \theta d \theta - \int_{pi/2}^\pi P_l(cos \theta)sin \theta d \theta)[/tex]
and let x=cosØ
 
  • #3
Hi, thank you for the answer! i was thinking about this kind of solution but the reasoning of the text (that's probably what i got wrong) seems to suggest that since:
Code:
[tex] \sigma\_0 (cos\theta\) = constant = \sigma\_0 P_0 (cos\theta\) [tex]

(of course that's for theta between 0 and pi/2, otherwise it's the same, but negative constant)
Then all coefficients are zero except for l = 0, but then the two integrals just reduce to:

Code:
[tex] A_l =\sigma\_0 /\epsilon\_0 [tex]
Now if I'm not supposed to kill off the polynomials, i guess i just end up with the whole infinite series, for which the formula is straightforward...
(take you for the LaTex link, too, i hope i used it right..)
 

Related to What Is the Method for Finding the Potential of a Spherical Dipole?

1. What is a spherical dipole?

A spherical dipole is a type of electric or magnetic dipole, which is a pair of opposite charges or magnetic poles separated by a short distance. In a spherical dipole, the charges or poles are located on opposite sides of a spherical surface, rather than on a straight line as in a traditional dipole.

2. How is the potential of a spherical dipole calculated?

The potential of a spherical dipole can be calculated using the formula V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the magnitude of the charge, and r is the distance from the center of the dipole to the point of measurement. This formula assumes that the dipole is located in a vacuum, and the potential at any point is determined by the combined effects of the two charges or poles.

3. What is the significance of the potential of a spherical dipole?

The potential of a spherical dipole is significant because it can be used to understand the behavior of electric or magnetic fields around the dipole. By analyzing the potential, scientists can determine the strength and direction of the fields and how they will interact with other objects or particles in the vicinity of the dipole.

4. How does the potential of a spherical dipole change with distance?

The potential of a spherical dipole follows an inverse relationship with distance, meaning that as the distance from the dipole increases, the potential decreases. This relationship is described by the inverse square law, which states that the potential is inversely proportional to the square of the distance from the dipole.

5. Can the potential of a spherical dipole be manipulated?

Yes, the potential of a spherical dipole can be manipulated by changing the magnitude or distribution of the charges or poles. This can be achieved by physically moving the dipole or by adjusting the strength of the charges or poles. Additionally, external electric or magnetic fields can also influence the potential of a spherical dipole.

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