2 questions in electromagnetism.

1. Oct 27, 2009

MathematicalPhysicist

1. The problem statement, all variables and given/known data
1. In a dilectric sphere with radius a and polarization vector P= P0 r (r is the spherical radial vector) where P0>0. find D,E and the polarization charge volume density and area density.
2. A chagrge q is displaced at the centre of a hollow dilectric sphere (with a,b as its internal an outer radii), the dilectric constant e(r) depends on the radial distance, find E,D,P.

2. Relevant equations
Maxwell (electrostatic) Equations and boundary conditions.

3. The attempt at a solution
For 1, I found the charge densities, $$\rho ' =-\nabla \b{P}= -3P0$$ and $$\sigma ' = -(\b{P}_{inside \ sphere} - \b{P}_{outside \ sphere})\cdot \b{n}= P_0 a$$, Now I don't know how to find D and E, from the boundary conditions I know that:
div D= $$4\pi \rho_{free}$$ and that $$\sigma_{free} = -(\b{D}_{inside \ sphere} - \b{D}_{outside \ sphere})$$, and E I can find after I find D, by the fact that D=E+4pi P.
I suppose that $$\sigma_{free}=-\sigma '$$ cause the net charge is zero, and the same with volume densities, from both equations and from the symmetry of the problem (D is spherical radial) I can find D, I am not sure if this is valid.

For 2, not sure either, if I find D then E=D/e(r), and P=(e(r)-1)/4pi)E, the question is how do I find D?

Last edited: Oct 27, 2009
2. Oct 27, 2009

jdwood983

Do you know the relations?

$$\mathbf{P}=\varepsilon_0\chi_e\mathbf{E}$$

and

$$\mathbf{D}=\varepsilon_0\mathbf{E}+\mathbf{P}$$

these should help with both problems

3. Oct 27, 2009

MathematicalPhysicist

Yes, I know them (but in cgs system).
OK, I see now how I can solve question #1, but not sure how to answer question #2, can you help me with question #2?

thanks.

4. Oct 27, 2009

MathematicalPhysicist

Wait a minute, the first equation is valid only for linear materials, in the first question it's not given that the material is linear (i.e E is proportinal to D).

5. Oct 27, 2009

jdwood983

It doesn't say it's not linear either. For nonlinear polarization,

$$P_i=\sum_j\varepsilon_0\chi_{ij}E_j$$

I would assume you should go with what I wrote first, since you don't have the matrices needed to solve the nonlinear term.

6. Oct 27, 2009

jdwood983

For problem 2, wouldn't this just be Gauss's Law in dielectric form?

$$\oint \mathbf{D}\,d\mathbf{S}=Q_{enc}$$

and from D, find E and P?

7. Oct 27, 2009

gabbagabbahey

It also doesn't say there aren't a bunch of free charges scattered about, but I see no reason to assume there are.

That still looks like a linear relationship to me (If you double E, you double P)....Just because the material doesn't polarize parallel to E, doesn't mean its nonlinear.

8. Oct 27, 2009

jdwood983

True, but in non-linear cases, it is not $$\mathbf{E}$$ that is the non-linear term, it's the $$\chi_{ij}$$ that is the non-linear term.

9. Oct 27, 2009

gabbagabbahey

That looks a little odd to me...are you sure the problem doesn't specify $\textbf{P}=P_0\hat{\textbf{r}}$ (i.e. the polarization has a constant magnitude and points radially outward)?

What makes you think the net charge s zero? Unless you are given information to the contrary, I would assume there are no free charges. You can then calculate E directly from the bound charges using Coulomb's Law (or Gauss' Law) and D can be determined directly from its definition.

Last edited: Oct 27, 2009
10. Oct 27, 2009

gabbagabbahey

I'm not sure what you mean by this....unless $\chi_{ij}$ are functions of $E$ (which is not at all implied by the way you've written the equation), the relationship is linear.

11. Oct 27, 2009

jdwood983

Normally you would Taylor expand the relationship into more terms:

$$P_i=\varepsilon_0\left(\sum_j\chi^{(1)}_{ij}E_j+\sum_{jk}\chi^{(2)}_{ijk}E_jE_k+\cdots\right)$$

where $$\chi^{(1)}$$ is your normal susceptibility and $$\chi^{(2)}$$ comes from the Pockels effect and so on, but I didn't feel like expanding it out further than I did because I doubt that the professor would have assigned such an problem (that is, I still assume it is linear).

So I guess I was technically wrong about saying that $$\chi_{ij}$$ was the non-linear term. When you asked further, I pulled out my non-linear optics text to double check.

12. Oct 27, 2009

MathematicalPhysicist

I am sure.
Why not, I use the nabla operator in spherical coordinates which according to my textbook is:
div A= 1/r^2 d/dr(r^2 A_r)

In this case, A_r=P0*r.

[/quote]

Not sure how to do this, $$div E=4\pi \rho '$$, I can assume here that E is radial, if yes then it's rather simple as well.

13. Oct 27, 2009

gabbagabbahey

Okay.

Yes, I edited my post when I realized my error

Have you not already calculated the fields due to a uniformly charged sphere and a uniformly charged spherical surface?

14. Oct 27, 2009

MathematicalPhysicist

So for r<a, Q enclosed is q, so D *4pi *r^2=q, => D=q/(4pi*r^2)
for a<r<b, D(4pi(r^2-a^2))=q
for r>b, D(4pi * b^2)=q

Is this right?

P.s
Don't know but electromagnetism in university is still my tough part in physics, though the theory is pretty neat I find myself baffled at questions which are rudimintary.

15. Oct 27, 2009

gabbagabbahey

Why do you have the r^2-a^2 there? What are you using as your Gaussian surface? What is the flux of a spherically symmetric D-field through that surface?

16. Oct 28, 2009

MathematicalPhysicist

I use the ring between the spheres with radius a and b, which its surface area is 4pi (r^2-a^2).

17. Oct 28, 2009

gabbagabbahey

I'm not sure what you mean by "ring", but I guess you are referring to a spherical shell of thickness r-a? If so, that is a VERY poor choice of Gaussian Surfaces; it is actually two surfaces (at r'=a and r'=r) and hence the D-field will have two different values over this surface (D(r'=a) and D(r'=r))

$$\oint\textbf{D}\cdot d\textbf{a}=4\pi D(r)r^2-4\pi D(a)a^2$$

Just use an infinitesimally thin spherical shell at r'=r instead.

18. Oct 28, 2009

MathematicalPhysicist

I still don't get it.

So, S D.da= 4pi D(r) r^2, or something else?

19. Oct 28, 2009

gabbagabbahey

Yes, so $\textbf{D}=\frac{q}{4\pi r^2}\hat{\textbf{r}}$ everywhere....