Calculating Dipole Moment and Potential for Spherical Shell Charge Distribution?

[minimark1234]

Homework Statement

find:

\vec{p} (dipole moment)
\itshape V(r,\theta) (dipole potential)

of a spherical shell which has a charge area of \itshape \sigma= k cos \theta.

Homework Equations

\itshape \vec{p}= \int\int \vec{r} \sigma (r\stackrel{2}{} sin \theta d\theta d\phi)
\itshape V(r, \theta )= \frac{\hat{R} \bullet \vec{p}}{4 \pi \epsilon R^{2}}

The Attempt at a Solution

I came up with 0 both times and i know that's wrong. please help

 

[vela]

Show us your work so we can see where you went wrong.

 

[jdwood983]

I can tell you that either your integral is wrong or you are not accounting for \vec{r} in the integral. Since I know the equation is correct, you are not accounting for \vec{r} in your calculation.

Do you know what \vec{r} is?

 

[minimark1234]

Work: OK i re-did the problem this time taking a different approach...

\itshape\vec{p}=\int^{\pi}_{0}\int^{2\pi}_{0}\vec{r}\sigma\:d\vec{a}

\itshape \sigma=kcos\:\theta\:,\:d\vec{a}=r^{2}sin\:\theta\:d\theta\:d\phi\:\hat{r}\:,\:k=constant

so...
\itshape \vec{p}=\int^{\pi}_{0}\int^{2\pi}_{0}r^{3}k(cos^{2}\:\theta\:sin\:\theta\:sin\:\phi\:\hat{x}\:+\:sin^{2}\:\theta\:cos\:\theta\:cos\:\phi\:\hat{y}\:+\:cos^{2}\:\theta\:sin\:\theta\:\hat{z})\:d\theta\:\:d\phi
so when...
\itshape R\geq\:r\;\Rightarrow\;\vec{p}=\frac{4}{3}\pi r^{3}k\:\hat{z}
and when...
\itshape r\geq\:R\;\Rightarrow\;\vec{p}=\frac{4}{3}\pi R^{3}k\:\hat{z}

and finally...
\itshape \hat{r}\bullet\vec{p}=cos\:\theta\:\hat{z}\bullet\frac{4}{3}\pi r(or\:R)^{3}k\:\hat{z}=\frac{4}{3}\pi r(or\:R)^{3}k\:cos\:\theta\:

so when...
\itshaper R\geq\:r\;\Rightarrow\;V_{dip}(r,\theta)=\frac{kR^{3}}{3\epsilon r^{2}}\:cos\:\theta
and when...
\itshape r\geq\:R\;\Rightarrow\;V_{dip}(r,\theta)=\frac{kr}{3\epsilon}\:cos\:\theta
but what disturbs me is i thought i was supposed to be finding an approximate amount of V... and that was the exact answer, did i go wrong somewhere and just get really lucky giving me the exact amount, or am i right... and if I am right what does using a higher multipole change?

 

[minimark1234]

@jdwood983 the first time i didn't account for \itshape \vec{r}=r\hat{r}, and i always forget to because I'm being lazy, the second time i did though.

 

[Phyisab****]

Is it not possible that this charge distribution has exactly the electric field of a certain dipole? In which case higher order terms would not be necessary.

 

[minimark1234]

I am not sure, nor do i know a way to check that. Is the a formula or and of the given i had above that would tell me this, I'm self teaching Griffiths' book, which has no answers and no teacher to ask, so I'm a bit lost.

 

[Phyisab****]

It is certainly possible, and your result seems like more than a coincidence given what he asks in part b.

 

[minimark1234]

Because what really drove the integral above was the \itshape \sigma which says that the charge density on the x and y-axis are both k (when z=0) would it be safe to say the the dipole moment will be in the z direction in cases like this? Also because of circle symmetry the net charge is 0 at the origin (I assume is the center) is the V going to be defined as a dipole, which i think might be mentioned in the book?

 

[vela]

It's been too long since I've done this stuff, so I might be totally off here. But I'd guess the result isn't too surprising because \sigma \propto Y^0_1, where Y^m_l are the spherical harmonics.

 

[minimark1234]

the whole Y thing went right past me, but what u said i think clarified my final troubles on this problem. Thank you everyone for the help on this problem :)