1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Griffiths page 176

  1. Mar 16, 2008 #1
    [SOLVED] Griffiths page 176

    1. The problem statement, all variables and given/known data
    Below example 4.4, Griffiths says "We cannot apply Gauss's law precisely at the surface of a dielectric, for rho_b blows up..."

    Why does rho_b blow up?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 17, 2008 #2
  4. Mar 17, 2008 #3


    User Avatar
    Science Advisor

    That sentence is a bit convoluted, but G is probably referring to bound surface charge density, which corresponds to infinite volume charge density.
    If a discussion in GIII confuses you, go back to GI, which has less of that confusing help.
  5. Mar 17, 2008 #4
    Hello ehrenfest,it is an interesting question and I wonder if I myself understood this point while I read this section a year or two ago.

    The problem is with the definition of [tex]\rho_b=\ - \ div \ P [/tex] and [tex]\sigma_b=\ P.n[/tex].UNDERSTAND that P is determined by [tex]\sigma_b[/tex] and [tex]\rho_b[/tex]

    Note that when we say there is a quantity [tex]\rho_b[/tex] and another quantity [tex]\sigma_b[/tex],we generally do not care if there is any sharp boundary between the charge distributions.I mean we do not say upto 0.005 cm below the surface we would call [tex]\sigma_b[/tex] and below that we would call as [tex]\rho_b[/tex]

    As Griffiths gives hint the charge density is expected to be gradually fade out toward the boundary.This makes [tex]\rho_b[/tex] a continuous function and hence P is also continuous.POLARIZATION GRADUALLY TAPERS OFF TO ZERO.This exculdes the existence of [tex]\sigma_b[/tex]

    But if there were a sharp boundary like I mentioned, polarization P would be discontinuos across that boundary near (on) the surface and by taking the divergence you will get [tex]\rho_b=\infty[/tex].

    I hope the problem is now clear to you.
  6. Mar 17, 2008 #5
    I see. So you are saying that a uniformly polarized object of finite size is unphysical.
    Last edited: Mar 17, 2008
  7. Mar 17, 2008 #6

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Not quite. What he's referring to is that any surface charge distribution can be written as a volume charge distribution by means of delta functions. For example, the surface charge density of a uniformly charged sphere can be written

    [tex]\rho(r, \theta, \phi) = \frac{Q}{4\pi a^2}\delta(r - a)[/tex]

    Naturally, this function blows up on the surface itself, just as the charge density (and divergence of E) blows up at a point charge.
  8. Mar 17, 2008 #7
    I meant not exactly this,but like this.Refer to Griffiths problem 1.45 (b).The derivative of a step function is a delta function.Likewise If there is a distinguishable sharp boundary like between core and cover (jacket) of a body,rho_b must blow up.

    In reality you might have a uniform polarized object.But do not be too obsessed with the distinction of rho_b and sigma_b---they are the different faces of the SAME distribution and is not different like core and cover (jacket).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook