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E and D fields of polarizable material

  • Thread starter thefireman
  • Start date
  • #1

Homework Statement


A semi infinite slab of material, [tex]-\infty<x<0, -\infty<z<\infty, -d/2<y<d/2[/tex]
has uniform polarization P in the +y direction.

What are the E and D field along x axis at y=0.

Homework Equations


[tex]\oint \vec{D}\cdot d\vec{a} = Q_{free enclosed}[/tex]
[tex]D= \epsilon_{0}E+P[/tex]

The Attempt at a Solution


If I use the integral statement, there is no free charge, D is uniformly 0.
This seems uninteresting and incorrect...as well as inconsistent with the second relation.

I already know that at the x=y=0 on the boundary, the magnitude of E field drops to 1/2 its "ideal infinite capacitor" (from surface charge due to polarization). IF D were always 0 and P is uniform, than E would have to be uniform, which it is not.

So why is the integral invalid? I recall qualitatively, that the E and D field are inversely related within the material i.e. if one increased the other decreased...outside trivially they are equal, with a dielectric multiplication. How do I correctly calculate D?

Thanks in advance

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6

The Attempt at a Solution


If I use the integral statement, there is no free charge, D is uniformly 0.
This seems uninteresting and incorrect...as well as inconsistent with the second relation.

I already know that at the x=y=0 on the boundary, the magnitude of E field drops to 1/2 its "ideal infinite capacitor" (from surface charge due to polarization). IF D were always 0 and P is uniform, than E would have to be uniform, which it is not.

So why is the integral invalid? I recall qualitatively, that the E and D field are inversely related within the material i.e. if one increased the other decreased...outside trivially they are equal, with a dielectric multiplication. How do I correctly calculate D?

Thanks in advance
The integral is always valid, but not always useful: how do you know that D is uniform and exhibits planar symmetry?

In vacuum, you can tell the symmetries of E by looking at the charge distribution....if the charge distribution is symmetrical, so is E. The same does not hold true for D. The reason is that while E is always completely determined by its divergence (which is proportional to the charge) D is not. Because unlike curl(E), curl(D) is not always zero....I'll leave it to you to determine where curl(D)=curl(P) does not equal zero in this case....:wink:

The symmetries of E are always determined (at least in electrostatics) by the symmetries of the charge distribution, but the symmetries of D are only determined by the symmetries of the free charge distribution in cases where curl(D)=0 everywhere.
 
Last edited:
  • #3
Ok, i knew that fact, but had forgotten that setting integral to Qenclosed required assuming information about D. Thanks.
One final question. I need to relate the E field outside this faux capacitor to that inside...specifically,
Find a relationship between the electric fields in the y = 0 plane at x = a and
x = -a.
If I do a line integral E dl around a square bounded by x=-a and x=a, at symmetric y values, the y line integrals should cancel by symmetry, the total the sum should be 0... essentially the E field outside at x=a should be negative that inside at x=-a? This doesnt make sense to me, an infinite fringe field outside? shouldnt it die off the further we go into vaccum?

thanks again!
 

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