1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electromag question:

  1. Apr 28, 2005 #1
    I've been stuck on this for a while now and it has been frustrating me to the point of pain. Any help would be much appreciated..

    I'm 3rd year undergrad level by the way.. See attached:

    Attached Files:

  2. jcsd
  3. Apr 29, 2005 #2


    User Avatar
    Homework Helper

    For part a) derive a linear relationship between P and D... perhaps you already have the relation derived in your text. P=kD (find out what k should be)

    del.D=true charge density
    del.P=bound volume charge density

    so del.(kD)=bound charge density
    k del.D = bound charge density
    k(true charge density) = bound volume charge density.

    You can get the true charge density using charge/volume of layer. Then you can get the bound volume charge density using the above equation.

    For part b), use Gauss law for P using the two faces of the dielectric... it should be easy after getting part a). You can simply calculate the total bound charge using the answer of a) times the volume of the layer. The value of P at the face is the bound surface charge density.
    Last edited: Apr 29, 2005
  4. Apr 29, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure I can help you as well as some others might. But nobody else has jumped in here, so let me give it a try. I was never too fond of those cgs units, but that seems to be where this is coming from. In those units.

    [tex]\frac{P}{E} = \frac{\epsilon-1}{4\pi}=\frac{2.2}{4\pi}[/tex]

    if I am interpreting the given information correctly.

    [tex]P} = \frac{\epsilon-1}{4\pi}=\frac{2.2E}{4\pi}=\frac{D-E}{4\pi}[/tex]

    Assuming all the embedded charge is confined to the d = 2mm strip in the middle of the material, the electric field from that charge at the surfaces will be found from Gauss's law

    [tex](E_2 - E_1 ) \bullet \hat n = 4\pi \rho d[/tex]

    where [itex]\rho[/itex] is the average charge density within the strip accross the thinckness d.

    I don't think there is any contribution here from polarization charge, because the dielectric continues throughout the excess charge region, and is electrically neutral.

    If you find E, you can find P, which is proportional to E, and if you can find P you can find the bound surface charge.

    Now if you look deeper inside the 2mm layer, and assume a uniform excess charge density within that layer, the electric field is going to vary because the charge contained within a thinner strip is going to vary. The field in the center of the strip will be zero and will increase as you move out to the surface of the excess charge layer.

    There will be a polarization charge layer at the outside surface of the dielectric that will yield a bound surface charge density out there

    Check up on me wrt units and such, but I think this is basically the path you need to follow. If any of this makes sense, see what you can do with it. If you post some work on it, someone who is fresher than me on this stuff might jump in, and if not I'll look at it harder.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electromag question:
  1. Electromag problem (Replies: 4)

  2. Electromag problem (Replies: 1)