Quadrupole term for uniformly charged sphere (where did I go wrong?)

In summary, the conversation discusses an assignment that was returned with fewer marks than expected and confusion about a particular part. The professor is only available for a tutorial on Monday, but the person would like to figure out the mistake before then. They ask if anyone can spot the error and for the person to explain their reasoning using LaTex. The person eventually realizes their mistake and corrects it. The correct Legendre polynomial is given for ##n=2##.
  • #1
snatchingthepi
148
38
Homework Statement
Derive the quadrupole term for a general 3d potential
Relevant Equations
Multiple expansion
So I got an assignment returned to me with fewer marks than I had expected. One part in particular is confusing to me. The professor is only available on Monday for a tutorial, but I'd like to see what is wrong before then.

Can anyone spot why this is incorrect?
 

Attachments

  • 15690090530587869854179905602916.jpg
    15690090530587869854179905602916.jpg
    60.2 KB · Views: 209
  • 15690091265761269373936664742371.jpg
    15690091265761269373936664742371.jpg
    49.2 KB · Views: 220
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
Can you write what you have done, using LaTex, and explain what your reasoning is at each step?
 
  • Like
Likes Cryo and Delta2
  • #3
Actually I think I may have found the mistake as I confused a part of my notation with something that should have been unprimed. Thanks.
 
  • #4
You have also made a mistake in writing
##P_2(\cos\theta ')=\dfrac{3}{2}\cos\theta '-\dfrac{1}{2}##.
The correct Legendre polynomial for ##n=2## is
##P_2(\cos\theta ')=\dfrac{3}{2}\cos^2\theta '-\dfrac{1}{2}##.
 
  • Like
Likes snatchingthepi
  • #5
kuruman said:
You have also made a mistake in writing
##P_2(\cos\theta ')=\dfrac{3}{2}\cos\theta '-\dfrac{1}{2}##.
The correct Legendre polynomial for ##n=2## is
##P_2(\cos\theta ')=\dfrac{3}{2}\cos^2\theta '-\dfrac{1}{2}##.
Thank you!
 

Related to Quadrupole term for uniformly charged sphere (where did I go wrong?)

1. What is the quadrupole term for a uniformly charged sphere?

The quadrupole term for a uniformly charged sphere is a mathematical representation of the second-order moment of the charge distribution of the sphere. It takes into account the deviation of the charge distribution from a perfect sphere, which results in a non-zero quadrupole moment.

2. How is the quadrupole term calculated for a uniformly charged sphere?

The quadrupole term can be calculated using the formula Q = ∫(3cos²θ - 1)dV, where Q is the quadrupole moment, θ is the angle between the direction of the charge and the axis of symmetry, and dV is the volume element.

3. Why is the quadrupole term important in studying charge distributions?

The quadrupole term provides a more accurate representation of the charge distribution compared to just the monopole term (charge at the center) and dipole term (charge separation along an axis). It helps in understanding the shape and symmetry of the charge distribution, which is crucial in various fields such as electrostatics and magnetostatics.

4. Can the quadrupole term for a uniformly charged sphere be zero?

Yes, it is possible for the quadrupole term to be zero if the charge distribution is perfectly spherical. In this case, the charge distribution does not deviate from the ideal spherical shape, resulting in a zero quadrupole moment.

5. Where could I have gone wrong in calculating the quadrupole term for a uniformly charged sphere?

There are several factors that could lead to an incorrect calculation of the quadrupole term, such as using the wrong formula or not considering the correct boundary conditions. It is important to carefully check all the steps and assumptions made in the calculation to ensure accuracy.

Similar threads

Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
7K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
2K
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Back
Top