Potnetial of a spherical Shell (1 Viewer)

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

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

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

and the radius of the sphere is constant theta and phi are constant
so that
[tex] p = \int_{0}^{\pi} \int_{0}^{2 pi} \theta' \sigma(\theta') R^2 \sin\theta' d \theta' d \phi [/tex]
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:

quasar987

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Look at equ. 3.98. p and r' are VECTORS in there.
 
I take it you are refering 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 cant use theta the way i used it

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

but the integral
[tex] \int_{0}^{2\pi} \cos\theta' \sin\theta' d\theta = 0 [/tex]!
cant have zero dipole moment...
 
HalfManHalfAmazing said:
I take it you are refering to Griffith's textbook?
problem 3.28
page 151
 

quasar987

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[tex]\vec{r}=r\hat{r}=r(\hat{x}\sin\theta \cos \phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta)[/tex]
 

quasar987

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

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