Electric field due to a point dipole

  • #1

Main Question or Discussion Point

Let's say I have a point dipole (as an approximation for an atom) at the origin and it oscillates in the [tex]z[/tex] axis. The (theta component of the) electric field due to this dipole in the far field will be

[tex]E = \frac{d}{4\pi\epsilon_0}\frac{k^2\sin\theta}{r}\exp i\left(kr-\omega t\right)[/tex]

where [tex]d[/tex] is the dipole moment, [tex]k=2\pi/\lambda[/tex], [tex]\theta[/tex] is the angle made with the [tex]z[/tex] axis, [tex]r[/tex] is the radial distance, [tex]\omega[/tex] is the angular frequency of the oscillation and [tex]t[/tex] is the time.

Due to the zero [tex]\phi[/tex] dependance, i.e. the angle in the equatorial plane, there is a cylindrical symmetry. Instinct tells me that I should I have two lobes of electric field, one in the [tex]+z[/tex] direction and one in [tex]-z[/tex], which will oscillate, alternatively between positive and negative. However the equation I quoted implies a doughnut shaped electric field. For a given [tex]r[/tex], [tex]E[/tex] increases as [tex]\theta[/tex] goes from 0 to [tex]\pi/2[/tex], then decreases from [tex]\pi/2[/tex] to [tex]\pi[/tex]. I.e. I think the equation should have a [tex]\cos\theta[/tex] in it instead of a [tex]\sin\theta[/tex]

Where have I gone wrong? I think it's my definition of [tex]\theta[/tex], however it is always defined from the [tex]z[/tex] axis.

PS. I'm pretty sure the equation is in Jackson.
 
Last edited:

Answers and Replies

  • #2
G01
Homework Helper
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This is the correct form for a point dipole oscillating along z. The intensity profile for a point dipole source is indeed like a donut.

Since the dipole is oscillating along z, there is 0 intensity at [itex]\theta=0[/itex], and the intensity is maximum [itex]\theta=\pi /2[/itex] in the x-y plane. I think if you take a cross section of the donut on the z-y or z-x planes you'll find the lobes you are picturing. However, the lobes are centered in the x-y plane, not along z.

For reference, see Griffith's Eq. 11.18 and Figure 11.4
 

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