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Homework Help: Griffiths E&M: Do bound charges physically exist?

  1. Mar 20, 2010 #1
    Well, actually it's not a homework problem, I just got really confused about bound charges.
    Originally, I thought it was just a special technique to do the integral, but somehow Griffiths suggest that bound charges are phsically exist.(chap. 4.2.2) Well, I can accept his argument, but later when we do the boundary condition problem, we have the equation
    Integral D dv = Q (free)
    So why it's Q free, not Q free plus the surface bound charge ? I know the math why the equation is true, but if surface bound charge physically exist, I don't think the equation make any sense...

    Any help is appreciated..
  2. jcsd
  3. Mar 20, 2010 #2
    The bound charges do exist. You see them in:

    [tex]\epsilon_0 \nabla \cdot \vec{E} = \rho = \rho_{free} + \rho_{bound}[/tex]

    Don't forget:

    [tex]\vec{D} = \epsilon_0 \vec{E} + \vec{P}[/tex]

    So the bound charges are absorbed in the polarization vector.
  4. Mar 20, 2010 #3


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    Homework Helper

    Hi Jin! :smile:

    I don't have a copy of Griffiths, but if D in this is the electric displacement field , then by definition only the free charge is involved.

    (For P, only the bound charge is involved, and for E the total charge is involved)
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