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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


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    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

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    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).
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