Dipole moment.

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From Griffiths, Problem 4.1
A sphere of radius R carries a polarization
[tex]
\textbf{P}=k\textbf{r}
[/tex]

where k is constant and r is the vector from the center.
a. Calculate [tex]\sigma_b[/tex] and [tex]\rho_b[/tex].
b. Find the field inside and outside the sphere.

part a is handled simply by [tex]\sigma_b=\textbf{P}\cdot\hat{n}[/tex] and [tex]\rho_b=-\nabla\cdot\textbf{P}[/tex].

part b is handled most easily by using the bound charges found and gauss's law, giving: [tex]\textbf{E}=\frac{-kr}{\epsilon_0}\hat{r}[/tex] and 0 outside.

part b can also be handled by first getting the potential through:[tex]\frac{1}{4\pi \epsilon_0}\int_v \frac{\hat{r}\cdot\textbf{P}}{r^2}d\tau[/tex]. Then just get the negative gradient of V to yield E. This gives the same result as method 1 for both inside and outside (P=0 when dealing with outside case).

Now, i tried to get the field, again by computing for V but this time with:
[tex]V(r)=\frac{1}{4\pi \epsilon_0}\oint_s\frac{\sigma_b}{r}da + \frac{1}{4\pi \epsilon_0}\int_v\frac{\rho_b}{r}d\tau[/tex]. The answer i get using this integral is different. i would like to know if in the first place this integral is applicable? if i'm trying to get the potential inside, do i still consider the surface bound charge?

the best i managed was an answer negative to the correct one. (but this had 1 dubioius step involved). also, i could only attempt to get the potential inside. can anyone make the last method work (for both outside and inside)? or share why it doesnt?

thanks.
 

Answers and Replies

  • #2
Meir Achuz
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\sigma_b is not spherically symmetric, so E iinside is not.
You can't use Gauss as you do in that case.
Your integrals should give the right answer if you mean r-r'
when you write r.
 
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  • #3
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Gauss's law actually gives the right answers. [tex]\sigma_b=kR[/tex] everywhere on the surface if you apply the definition. likewise, [tex]\rho_b=-3k[/tex]. these are symmetric and so is E inside and out.

as for the r-r' hint. i don't see it yet, but i will try it. thanks.

i was wrong though, about my claim that i got E=0 outside using the second method. that's another issue - how to get that result.
 
  • #4
Meir Achuz
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abode_x said:
Gauss's law actually gives the right answers. [tex]\sigma_b=kR[/tex] everywhere on the surface if you apply the definition. likewise, [tex]\rho_b=-3k[/tex]. these are symmetric and so is E inside and out.

as for the r-r' hint. i don't see it yet, but i will try it. thanks.

i was wrong though, about my claim that i got E=0 outside using the second method. that's another issue - how to get that result.
\sigma_b does not equal kR. It equals kR cos\theta because of the dot product. Go back and study Griffriths some more, then come back to the problem.
 
  • #5
siddharth
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Meir Achuz said:
\sigma_b does not equal kR. It equals kR cos\theta because of the dot product. Go back and study Griffriths some more, then come back to the problem.
Meir Achuz, I don't quite understand this.

If [itex]P(r)=kr \vec{e_r} [/itex]
Then, won't [tex]\sigma_b[/tex] be

[tex]\sigma_b = \vec{P}.\hat{n} = k R (\vec{e_r}.\vec{e_r}) = kR [/tex]
 
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  • #6
Meir Achuz
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Whoops. I apologize. I had read it as P=a fixed vector.
Using Gauss as in the first post gives the right answer.
To use the potential integrals still requires r-r'.
Thanks for catching my carelessness.
 

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