Potnetial of a spherical Shell

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The discussion focuses on calculating the dipole moment of a spherical shell with a surface charge density that varies with the angle, specifically σ = k cos(θ). The initial approach incorrectly treats θ as a constant, leading to confusion about the resulting dipole moment being negative. Participants clarify that both the position vector and the dipole moment are vectors, and emphasize the importance of correctly integrating over the spherical coordinates. It is noted that the integration of certain terms results in zero, which raises concerns about the physical interpretation of the dipole moment. The conversation highlights the need to carefully consider the vector nature of the variables involved in the calculations.
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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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