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Homework Help: Electric flux density of oscillating electric charge

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data

    The electric flux density in free space produced by an oscillating electric charge placed at the origin is given by
    [tex]\vec{D}=\hat{r}\frac{10^{-9}}{4\pi r^2}cos(wt-\beta r), \ \ where \ \beta=w \sqrt{\mu_0 \epsilon_0}[/tex]
    Find the time-average charge that produces this electric flux density

    2. Relevant equations
    [tex]div\vec{D}=\rho \ (1) \\ \int_S \vec{D}.\vec{ds}=Q_{int} \ (2)[/tex]

    3. The attempt at a solution
    This is the exercise 1.9 from Balanis' Advanced Engineering Electromagnetics. As the charge is oscillating, we can see a propagation delay in the cosine argument. Applying the divergent in D in spherical coordinates, we get the value
    [tex]div\vec{D}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 cos(wt-\beta r)\frac{10^{-9}}{4\pi r^2}) = \beta sin(wt-\beta r)\frac{10^{-9}}{4\pi r^2}[/tex]
    And that is the value of the charge density. As expected, it goes to infinity when r→0, but it isn't zero when r≠0. In the other hand, equation (2) gives the value of internal charge Qint=10-9cos(wt-βr), as expected from a oscillating punctual charge. Why equation (1) gives a wrong value?
    Last edited: Mar 8, 2016
  2. jcsd
  3. Mar 9, 2016 #2


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    Hello @Fernando Valadares,

    Welcome to PF! :welcome:

    I think you missed a minus sign in there. :wink: Other than that, it looks good to me so far.

    [Edit: Ooops. That minus sign was my mistake, not yours. :oops: Sorry about that. Your answer looks good as it is. :smile:]

    That also looks good to me. :smile:

    Besides the absence of the minus sign in your first answer, they both look correct to me.

    Perhaps you are confusing charge density with the charge itself.

    In order to determine the charge from the charge density, you must integrate the charge density over the volume. You cannot simply multiply times the volume of the Guassian surface in this case because the charge density is not uniform (it is dependent upon r).

    You really need to do the integration when doing this approach.

    [tex] Q_{int} = \iiint_V \rho \ dV [/tex]

    where [itex] dV [/itex] is the volume element and, depending on the variables you use for your spherical coordinate system, will be of the form:
    [itex] dV = r^2 \sin \theta \ dr \ d \theta \ d \phi [/itex]
    (I don't have Balanis' Advanced Engineering Electromagnetics, so I don't know what conventions the book uses. I'm just saying the volume element will look something like above, although it might use different variables or have variables swapped around. You should use what your book uses.)

    So first, fix that negative sign in charge density, [itex] \rho [/itex].

    Second, [well, first actually] evaluate that volume integral of the charge density and it should match your second answer. Good luck! :smile:
    Last edited: Mar 9, 2016
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