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Maxwell’s equations for oscillating electric dipole

  1. Apr 17, 2012 #1
    How do I show that our equations for the E- and B-fields for the oscillating electric dipole do NOT satisfy Maxwell’s equations?

    After approximations in retarded potentials, we have our E- and B-field as following:

    E = -ω2[itex]μ_{0}[/itex][itex]p_{0}[/itex](4∏r)-1sin(θ)cos[ω(t-[itex]\frac{r}{c}[/itex])][itex]\hat{θ}[/itex] (Griffiths 11.18)

    and

    B = -ω2[itex]μ_{0}[/itex][itex]p_{0}[/itex](4∏cr)-1sin(θ)cos[ω(t-[itex]\frac{r}{c}[/itex])][itex]\hat{\phi}[/itex] (Griffiths 11.19)

    Where ω is angular frequency for the oscillating charge moving back and forth, c is the speed of light, r is the distance where E and B are to be calculated, θ is the angle between dipole axis and the distance r, [itex]p_{0}[/itex] is the maximum value of dipole moment, [itex]μ_{0}[/itex] is permeability of free space, t is time, [itex]\hat{\phi}[/itex] is direction in azimuthal angle, and [itex]\hat{θ}[/itex] is direction in polar angle.

    I got divergence of B is satisfied (2nd eq. of Maxwell's), also, I got faradays law satisfied (3rd eq. with curl of E).

    I am stuck in the other two equations:

    For Gauss's law (1st eq.) I got div. of E does not equal zero, but maybe that because of the charge density. So, I am not sure whether this equation is satisfied or not, and I do not know how to show that.

    Also, the same argument For Curl of B. I got the same result for time derivative of E in addition to an extra component in [itex]\hat{r}[/itex] direction which may be the volume current density term in 4th Maxwell's equation (Ampere's and Maxwell's law).
     
  2. jcsd
  3. Apr 17, 2012 #2

    Bill_K

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    Hey, one reason these don't satisfy Maxwell's Equations is because they are not the correct fields. They are only approximations, valid at large distances. The field of an electric dipole, oscillating or not, contains 1/r2 and 1/r3 terms.
     
  4. Apr 17, 2012 #3
    So, does Maxwell's 1st and 4th equations reduces to [itex]\nabla[/itex].E = 0 and [itex]\nabla[/itex]×B = ε[itex]_{0}[/itex]μ[itex]_{0}[/itex][itex]\frac{\partial}{\partial t}[/itex]E in this case?
     
  5. Apr 17, 2012 #4

    DrDu

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    I wouldn't say so. If it contains such terms, it is not a pure dipole. Think of a sphere whose surface charge varies like cos theta.
    However, even the field of a point dipole contains an additional delta function contribution at r=0, see
    http://en.wikipedia.org/wiki/Dipole
     
  6. Apr 17, 2012 #5

    Born2bwire

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    If you calculate the fields for a point-source electric dipole you do get 1/r, 1/r^2 and 1/r^3 terms. Specifically, the fields in cylindrical coordinates for a VED become,

    [tex] E_\rho = \cos \theta \sin \theta \left( \frac{3}{r^2} - \frac{i3k}{r} - k^2 \right) \frac{e^{ikr}}{r} [/tex]
    [tex] E_z= \left[ k^2\sin^2\theta - \frac{1}{r^2} + \frac{ik}{r} + \cos^2\theta \left( \frac{3}{r^2} - \frac{i3k}{r} \right) \right] \frac{e^{ikr}}{r} [/tex]

    So the \theta component in spherical coordinates becomes

    [tex] E_\theta = \cos\theta E_\rho - \sin \theta E_z = \sin \theta \left( -k^2 - \frac{ik}{r} + \frac{1}{r^2} \right) \frac{e^{ikr}}{r} [/tex]

    where k is the wavenumber and we assume an [itex]e^{-i\omega t}[/itex] time dependence. These are derived via Maxwell's Equations. So the first-order term of the E_\theta component still lines up with what the OP was given by Griffiths. The k^2 coefficient gives us the \omega^2 and taking the real part would give use the cosine component.

    But in terms of the OP's question, the divergence of the electric field should evaluate to zero. Technically, there are sources at the origin, but since this is an oscillating dipole then the source is not a charge (hence why the divergence of the electric field is zero) but a current. So, everywhere but the origin you should find:

    [tex] \nabla\cdot \mathbf{E} = 0 [/tex]
    [tex] \nabla \times \mathbf{B} = \mu_0\epsilon_0 \frac{\partial \mathbf{E} }{\partial t} [/tex]
     
    Last edited: Apr 17, 2012
  7. Apr 17, 2012 #6

    DrDu

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    You are obviously right. I only wanted to say that the field of a non oscillating dipole does not contain a 1/r2 term.
     
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