Potnetial of a spherical Shell

[stunner5000pt]

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 :)

 

[quasar987]

Look at equ. 3.98. p and r' are VECTORS in there.

 

[HalfManHalfAmazing]

I take it you are referring to Griffith's textbook?

 

[stunner5000pt]

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...

 

[stunner5000pt]

I take it you are referring to Griffith's textbook?

problem 3.28
page 151

 

[quasar987]

$$\vec{r}=r\hat{r}=r(\hat{x}\sin\theta \cos \phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\theta)$$

 

[quasar987]

$\hat{r}$ 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 $\hat{x},\hat{y},\hat{z}$.