Potnetial of a spherical Shell

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SUMMARY

The discussion focuses on calculating the dipole moment of a spherical shell with a surface charge density defined as \(\sigma = k \cos \theta\), as presented in Griffith's textbook problem 3.28. The user initially attempts to express the dipole moment using spherical coordinates but encounters issues with the integral yielding a zero result. Key insights reveal that the dipole moment \(\vec{p}\) must be treated as a vector, and the integration must account for the variable nature of the unit vector \(\hat{r}\) during the summation process. The correct formulation involves integrating \(\vec{r'}\) and \(\sigma(\theta')\) over the appropriate spherical coordinates.

PREREQUISITES
  • Understanding of Griffith's Electrodynamics, specifically problem 3.28
  • Familiarity with vector calculus in spherical coordinates
  • Knowledge of dipole moments and their physical significance
  • Proficiency in performing double integrals over spherical surfaces
NEXT STEPS
  • Review Griffith's Electrodynamics, focusing on Chapter 3 and problem 3.28
  • Study vector calculus, particularly the treatment of unit vectors in integrals
  • Learn about the physical interpretation of dipole moments in electrostatics
  • Practice solving similar problems involving spherical charge distributions
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking to clarify concepts related to dipole moments and spherical charge distributions.

stunner5000pt
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Griffith's EM problem 3.28
A spherical shell of radius R has a surface charge \sigma = k \cos \theta

a) Calculate the dipole moment of this charge distribution.
i know that
p = r' \sigma(r') da'

but here sigma depends on theta
would the dipole moment p then turn into
p = \theta' \sigma(theta') da'

and the radius of the sphere is constant theta and phi are constant
so that
p = \int_{0}^{\pi} \int_{0}^{2 pi} \theta' \sigma(\theta') R^2 \sin\theta' d \theta' d \phi
i get a negative dipole moemnt as a result of this though... which amkes no sense
what am i doing wrong??

please help!

thanks :)
 
Last edited:
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Look at equ. 3.98. p and r' are VECTORS in there.
 
I take it you are referring to Griffith's textbook?
 
quasar987 said:
Look at equ. 3.98. p and r' are VECTORS in there.

right they are vectors...

so then i can't use theta the way i used it

so
\vec{p} = \int \vec{r'} \simga(\theta') d\vec{a'}
p = \int_{0}^{\pi} \int_{0}^{2\pi} r' k \cos\theta' r'^2 \sin\theta d\theta d\phi

but the integral
\int_{0}^{2\pi} \cos\theta' \sin\theta' d\theta = 0!
cant have zero dipole moment...
 
HalfManHalfAmazing said:
I take it you are referring to Griffith's textbook?

problem 3.28
page 151
 
\vec{r}=r\hat{r}=r(\hat{x}\sin\theta \cos \phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta)
 
\hat{r} is not a vector like \hat{x},\hat{y},\hat{z}. The latest are constants vectors while \hat{r} points towards the point that you're integrating (if I may say so). So it changes as you "sum" each d\theta and d\phi (if I may be so ruthless). So you can't pull it out of the integral as opposed to "inert" vectors like \hat{x},\hat{y},\hat{z}.
 

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