Multipole approximation outside conducting sphere

In summary: The grounded case is just a dipole with image charges pointing in the same direction, so the answer is the same for both cases. However, the neutral sphere case has an additional complication in that there are three charges at the sphere's surface: Q, the dipole moment, and -Q. The distance between Q and -Q is twice the distance between -Q and Q. This causes a moment of charge, μ, at the surface of the sphere. μ is equal to ##\vec{p}## times the area of the sphere.
  • #1
usn7564
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Homework Statement


A dipole is placed next to a sphere (see image), at a large distance what is E proportional to?

3. The Attempt at a Solution or lack thereof
I'm having trouble figuring out what's happening in any variations of these. How does the dipole affect the sphere's charge layout? Scavenging through Griffith's but not been able to find this in there and am at quite a loss.
 

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  • #2
usn7564 said:

Homework Statement


A dipole is placed next to a sphere (see image), at a large distance what is E proportional to?

3. The Attempt at a Solution or lack thereof
I'm having trouble figuring out what's happening in any variations of these. How does the dipole affect the sphere's charge layout? Scavenging through Griffith's but not been able to find this in there and am at quite a loss.
Here is a link to Wikipedia.
 
  • #3
SammyS said:
Here is a link to Wikipedia.

Isn't that just for a dipole inside the sphere?
 
  • #4
usn7564 said:
Isn't that just for a dipole inside the sphere?
No.

It is a bit strange to have the dipole inside. That simply means that the conducting sphere may be replaced by an image dipole outside the sphere.

If a dipole identical to the above image dipole is outside the sphere in the same location and orientation as the image dipole above, then there is an image dipole matching the above dipole inside the sphere.
 
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  • #5
SammyS said:
No.

It is a bit strange to have the dipole inside. That simply means that the conducting sphere may be replaced by an image dipole outside the sphere.

If a dipole identical to the above image dipole is outside the sphere in the same location and orientation as the image dipole above, then there is an image dipole matching the above dipole inside the sphere.
Oh right of course, thank you.
 
  • #6
Having some trouble getting any intuitive understanding from the wiki. What I was thinking (which may be way off) for the first image was that the spheres charges will rearrange so there are more positive charges on the right creating a dipole moment pointing in the same direction as the dipole. And E is proportional to a dipole in that case.

But the grounded case? No idea what's going on
 
  • #7
usn7564 said:
Having some trouble getting any intuitive understanding from the wiki. What I was thinking (which may be way off) for the first image was that the spheres charges will rearrange so there are more positive charges on the right creating a dipole moment pointing in the same direction as the dipole. And E is proportional to a dipole in that case.

But the grounded case? No idea what's going on
(Sorry for the long delay in responding.)It's clear that the answers to this exercise are qualitative rather than quantitative due to the limited information given.

I think the easiest case to handle is the following, the rightmost of the attached figures in the OP.

attachment.php?attachmentid=70171&d=1401471955.jpg


Unfortunately, the choice of variable names in that Wikipedia page aren't too appropriate when discussing a dipole with dipole moment of ##\vec{p}##, due to some conflicts in variable naming.Model the dipole as consisting of charges, Q & -Q arranged vertically and separated by some distance, Δ. Both of these are equal distance from the center of the sphere. Let q be the image charge for, Q. It will have a sign opposite that of Q and will be smaller in magnitude. Then the image charge for -Q is -q. The separation, δ, for the image charges is smaller than Δ. The resulting dipole formed by the image chrges is smaller than ##\vec{p}## and has opposite orientation.

How does that "look" from far far away ?
 
Last edited:
  • #8
usn7564 said:
Having some trouble getting any intuitive understanding from the wiki. What I was thinking (which may be way off) for the first image was that the spheres charges will rearrange so there are more positive charges on the right creating a dipole moment pointing in the same direction as the dipole. And E is proportional to a dipole in that case.

But the grounded case? No idea what's going on
The other two cases:
attachment.php?attachmentid=70169&d=1401471955.jpg
  . . .    
attachment.php?attachmentid=70170&d=14014719.jpg


The Wikipedia treatment of the image of a dipole can be helpful for these two cases, but I think it's also helpful to look at the dipole as a physical dipole with two charges, a positive charge, Q, and a negative charge, -Q, again separated by distance, Δ, the positive charge being farther from the sphere, the negative charge being closer.

Consider the locations, the signs, and the magnitudes of the resulting image charges.

The neutral sphere (non-grounded) case is actually a bit more complicated to analyze than the grounded case.
 

Related to Multipole approximation outside conducting sphere

1. What is the multipole approximation method?

The multipole approximation method is a mathematical technique used to simplify the analysis of the electric potential and field outside a conducting sphere. It involves expanding the potential and field in terms of spherical harmonics and using a finite number of terms to approximate the solution.

2. Why is the multipole approximation method used for conducting spheres?

The multipole approximation method is useful for conducting spheres because it allows for a simplified analysis of the electric potential and field outside the sphere. This method is particularly useful for larger spheres where the exact solution can be difficult to obtain.

3. How accurate is the multipole approximation method?

The accuracy of the multipole approximation method depends on the number of terms used in the expansion. As more terms are included, the accuracy of the approximation increases. However, for larger spheres, a small number of terms can still provide a reasonably accurate approximation.

4. Can the multipole approximation method be used for non-conducting spheres?

Yes, the multipole approximation method can also be used for non-conducting spheres, but the results may not be as accurate as for conducting spheres. This is because conducting spheres have a more well-defined boundary condition, making the approximation more accurate.

5. Are there any limitations to the multipole approximation method?

Yes, the multipole approximation method has limitations in its accuracy for highly non-uniform charge distributions and for points very close to the surface of the sphere. In these cases, a different method, such as numerical integration, may be more appropriate.

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