Why Use z Instead of r for Dipole Moment Calculation?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the dipole moment of a spherical shell with a specific charge distribution, σ = kcosθ, as presented in Griffith's electrodynamics. The original poster is exploring the use of different coordinate systems in the calculation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the definition of dipole moment using a spherical coordinate system but encounters issues with the integral yielding zero. They question the transition to using the z-component, Rcosθ, instead of r. Other participants suggest considering r as a vector and focusing on the symmetry of the charge distribution.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the vector nature of r and the implications of symmetry in the charge distribution. The original poster expresses a realization about the complexity of vector integration in spherical coordinates and seeks alternatives.

Contextual Notes

There is an indication of confusion regarding the treatment of the unit vector in spherical coordinates and the implications for the integral. The discussion reflects a learning process with varying levels of understanding among participants.

Idoubt
Messages
170
Reaction score
1

Homework Statement



I'm trying to do problem 3.28 in griffith's electrodynamics. The problem statement is, to find the dipole moment of a spherical shell with charge distribution σ = kcosθ

The way I tried to do it was to use the definition of dipole moment, which griffith defines as

P= ∫r σ dζ

where r = position of charge w.r.t origin ( in this case R ) and dζ is volume element.

The above integral gives 0 ( unless i did something stupid)

I looked up the solution manual and the way it does it is to use Rcosθ ie z instead of r. Can some1 explain why this is?
 
Physics news on Phys.org
You should consider r in the integral as a vector. and since the charge distribution has symmetry with respect to x and y axes, we only consider z component of r which is Rcosθ. You can check x and y and make sure that they are zero.
 
I see now. My problem was that even though I looked at it as a vector, I didn't realize that the unit vector [itex]\hat{r}[/itex] itself was a function of θ and [itex]\phi[/itex] and I took it out of the integral. When i rewrite it in cartesian co-ordinates and do the integral for each component ( cartesian unit vectors are constant so I can take it out of the integral ) it comes out fine. But when I look at this, vector integration with spherical co-ordinates seems very complicated, is there an easier way than rewriting in cartesian co-ords and integrating?
 
i got confused with this problem too, thank you for taking it out. :smile:
 
:smile:
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K