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Bound charges - Question about the maths. Did I calculate this correctly?

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    We know that a sphere of radius R carries a polarization P(r)=kr where k is a constant and r the vector from the centre. Calculate [tex]\sigma_b[/tex] and [tex]\rho_b[/tex]

    3. The attempt at a solution

    If we let the direction of polarization coincide with the z axis then:



    [tex]\rho_b=-\nabla kr\hat{r}[/tex]
    [tex]\rho_b=-kr (\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}) (\sin\theta\cos\phi\hat{x}+\sin\theta\sin\phi\hat{y}+\cos\theta\hat{z})[/tex]

    My query is mostly about these last three steps...are they correct? I'm still struggling to wrap my head around substituting the right infinitesimals for integration problems and, as in this case, the right coordinates for vectors.

    Thanks for the help!
  2. jcsd
  3. May 20, 2009 #2


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    If the polarization is radial, how can the direction of polarization coincide with the z axis?

    Technically a dot product:
    but you probably knew that ;-)

    Stop right there: you're mixing coordinate systems. Your derivative is expressed in Cartesian coordinates, but the radial vector is expressed in spherical coordinates. Now, you could keep going with this, by using the chain rule; you would have to calculate the coordinate derivatives
    [tex]\frac{\mathrm{d}r}{\mathrm{d}x}, \frac{\mathrm{d}r}{\mathrm{d}y}, \frac{\mathrm{d}r}{\mathrm{d}z}, \frac{\mathrm{d}\theta}{\mathrm{d}x}, \frac{\mathrm{d}\theta}{\mathrm{d}y}, \frac{\mathrm{d}\theta}{\mathrm{d}z}, \frac{\mathrm{d}\phi}{\mathrm{d}x}, \frac{\mathrm{d}\phi}{\mathrm{d}y}, \frac{\mathrm{d}\phi}{\mathrm{d}z}[/tex]
    but that'd be kind of a pain. It would be much easier to express the radial vector in terms of x, y, and z.
  4. May 21, 2009 #3
    The thing is, I don't KNOW if the polarization is radial...I'm basing my (obviously very dangerous) assumption on a similar example (there is only one, can you believe it) in my textbook. So what do you suggest? Is there a way around this?

    Sorry, LaTex oversight on my part :smile:

    Now you see, this is exactly my problem. I simply don't get this stuff :redface:

    I thought


    was expressing spherical coordinates as Cartesian :cry:

    So how do I start fixing this?
  5. May 21, 2009 #4


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    You are mixing the equations for the scalar components and the vector components.

    If you want the scalar x coordinate converted from spherical coordinates, then [tex]x=R\sin\theta\cos\phi[/tex].

    If you want the x component of a vector converted from spherical coordinates, then [tex]A_x = A_R\sin\thea\cos\phi+A_\theta\cos\thea\cos\phi-A_\phi\sin\phi[/tex].

    I always keep a set of photocopies of various conversion charts hung up in my cube for handy reference.

    But like diazona suggested, use the divergence operator that is for spherical coordinates, it will make this so much easier.
  6. May 21, 2009 #5
    Thanks for the input so far! :smile:

    OK, I'm still not sure if I fully understand this, but could you tell me if my revised attempt is at least on the right track?

    [tex]\rho_b=-\nabla kr\hat{r}[/tex]
    [tex]\rho_b=-k (\hat{r}\frac{d}{dr}+\hat{\theta}\frac{1}{3}\frac{d}{d\theta}+\hat{\phi}\frac{1}{r\sin\theta}\frac{d}{d\phi}) (r\hat{r})[/tex]

    I'm still not so sure if my vector expression [tex](r\hat{r})[/tex] is correct or if there's supposed to be a couple of theta's and phi's in there :frown:
  7. May 21, 2009 #6


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    The problem statement says, "P(r)=kr where k is a constant and r the vector from the centre". A vector from the center of the sphere is radial, by definition.

    You need the divergence, not the gradient:

    [tex]\rho_b=-\nabla \cdot \textbf{P}[/tex]

    Divergence in spherical coordinates

    You do have the correct form for P:

    Last edited: May 21, 2009
  8. May 21, 2009 #7


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    The equation for [itex]\rho_b[/itex] is not correct. It should be [itex]\rho_b=-\nabla \cdot \vec{P}[/itex]. Now look up the divergence in spherical coordinates and write down the polarization like this [itex]P(\vec{r})=kr \hat{r}[/itex].
  9. May 21, 2009 #8


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    Returning to the bound surface charge, you have


    As diazona noted, this should be a dot product:

    \sigma_b=\textbf{P} \cdot \hat{n}

    You have [itex]\textbf{P}=kr\hat{r}[/itex]. Now, what is [itex]\hat{n}[/itex] on the surface of a sphere, in spherical coordinates?
  10. May 21, 2009 #9
    *doh* :redface:


    Oh, ok...So I've been writing it as the dot product (divergence) but treating it as the gradient. Well done phyz! *sigh*

    [itex]\hat{n}[/itex] at the surface of a sphere would be perpendicular to the surface...mmmm...so wouldn't it simply be [itex]\hat{r}[/itex]?
  11. May 21, 2009 #10
    Let's try again...

    [tex]\sigma_b=\textbf{P} \cdot \hat{n}[/tex]
    [tex]\sigma_b=kr\hat{r} \cdot \hat{r}[/tex]

    [tex]\rho_b=-\nabla\cdot kr\hat{r}[/tex]
    [tex]\rho_b=-(\frac{1}{r^2}\frac{d}{dr}(r^2 kr))[/tex]

    If this is right, then I think I can finally say I'm beginning to understand this (fingers crossed) :biggrin:
  12. May 21, 2009 #11


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  13. May 21, 2009 #12


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    One more little thing... you have


    Lower-case [itex]r[/itex] is a variable. To get [itex]\sigma_b[/itex] at the surface of the sphere, you need to substitute the value of [itex]r[/itex] at the surface of the sphere, which is...

    Finally, as a "sanity check" for this type of problem, note that when you polarize an object, you don't add or subtract any charge from the object as a whole. So the total bound charge should be zero.
  14. May 22, 2009 #13
    Got it!

    Thanks for all the patience and help! Slowly but surely confusion is being beaten into submission with the big stick of comprehension :tongue2:

    Have an excellent day! :smile
    Last edited: May 22, 2009
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