Calculating Electric Field of a Polarized Sphere

In summary, the conversation discusses the problem of finding the bound charges and electric field inside and outside a sphere with a polarization \textbf{P}=k\textbf{r}. Calculations for \sigma_b and \rho_b using \textbf{P}\cdot\hat{n} and -\nabla\cdot\textbf{P} are provided. The use of Gauss's law and potential integrals to find the electric field are also discussed, with the conclusion that Gauss's law gives the correct result while the potential integrals require the use of r-r'. The conversation also addresses the issue of a dot product in calculating \sigma_b and clarifies that it should be \sigma_b=kR cos\theta.
  • #1
abode_x
11
0
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.
 
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  • #2
\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.
 
Last edited:
  • #3
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
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
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]
 
Last edited:
  • #6
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.
 

1. How do you calculate the electric field of a polarized sphere?

To calculate the electric field of a polarized sphere, you can use the formula E = k * P * (3cos^2θ - 1) / r^3, where k is the Coulomb constant, P is the polarization vector, θ is the angle between the polarization vector and the radius vector, and r is the distance from the center of the sphere to the point of interest.

2. What is a polarized sphere?

A polarized sphere is a sphere that has a non-uniform distribution of electric charges, resulting in a dipole moment. This means that the sphere has a positive and negative pole, with an electric field that points from the positive pole to the negative pole.

3. What is the significance of calculating the electric field of a polarized sphere?

Calculating the electric field of a polarized sphere is important in understanding the behavior and interactions of charged particles in the sphere's vicinity. It can also help in predicting the movement of particles and studying the properties of the electric field in different regions.

4. Can the electric field of a polarized sphere change?

Yes, the electric field of a polarized sphere can change depending on the orientation of the polarization vector and the distance from the sphere. It can also change if the sphere is subjected to external forces or if the distribution of charges within the sphere is altered.

5. Is there a difference between the electric field of a polarized sphere and that of a uniformly charged sphere?

Yes, there is a difference between the electric field of a polarized sphere and that of a uniformly charged sphere. A uniformly charged sphere has a constant electric field that points radially outward from the center, while a polarized sphere has a non-uniform electric field that varies with distance and direction due to the presence of a dipole moment.

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