Potnetial of a spherical Shell

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 :)
 
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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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