2 questions in electromagnetism.

[MathematicalPhysicist]

Homework Statement

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.

Homework Equations

Maxwell (electrostatic) Equations and boundary conditions.

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?

Thanks in advance.

 

[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

 

[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.

 

[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).

 

[jdwood983]

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).

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.

 

[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?

 

[gabbagabbahey]

It doesn't say it's not linear either.

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

For nonlinear polarization,

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

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.

 

[jdwood983]

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.

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.

 

[gabbagabbahey]

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.

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)?

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.

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.

 

[gabbagabbahey]

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.

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.

 

[jdwood983]

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.

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.

 

[MathematicalPhysicist]

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)?

I am sure.

Careful,

$$\rho'=-\mathbf{\nabla}\cdot\textbf{P}\neq-3P_0$$

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.

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.

[/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.

 

[gabbagabbahey]

I am sure.

Okay.

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.

Yes, I edited my post when I realized my error

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.

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

 

[MathematicalPhysicist]

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?

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?

Thanks in advance.

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.

 

[gabbagabbahey]

for a<r<b, D(4pi(r^2-a^2))=q

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?

 

[MathematicalPhysicist]

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

 

[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.

 

[MathematicalPhysicist]

I still don't get it.

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

 

[gabbagabbahey]

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