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Homework Help: Electric Field of a Polarized Sphere with Nonuniform Polarization

  1. Jul 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius R carries a polarization of [itex] \vec{P} = k \vec{r}[/itex], where k is a constant and [itex]\vec{r}[/itex] is the vector from the center.

    Find the bopund charges, and the field inside and outside the sphere

    2. Relevant equations
    [tex] - \nabla \cdot \vec{P} [/tex]
    [tex] \sigma _b = \vec{P} \cdot \hat{n} [/tex]
    [tex]V = \frac{1}{4 \pi \epsilon _{0}} \oint _{S} \frac{1}{r} \vec{P} \cdot d \vec{a} - \frac{1}{4 \pi \epsilon _{0}} \int _{V} \frac{1}{r}( \nabla \cdot \vec{P}) dV [/tex]

    3. The attempt at a solution

    Setting up the integral, I've used the displacement [itex] r = \sqrt{R^{2}+z^{2}-2Rz cos\theta}[/itex] by lining the observation point up with the z axis, and having the polarization [itex] \vec{P}[/itex] being orientated in any direction.

    [itex] \sigma _b = \vec{P}(\vec{r}) \cdot \hat{n}[/itex], because of the fact that P is now a vector function dependent on the r vector, this leaves me confused. I think this confusion stems from the meaning of the [itex]\vec{r}[/itex].The [itex]\hat{r}[/itex] stays constant at R, but the the [itex]\theta , \phi[/itex] change.

    And now on to the [itex]\rho _b = - \nabla \cdot \vec{P} [/itex]. In spherical: [tex]\nabla \cdot \vec{P}(\vec{r})= = \frac{1}{r^2} \frac{ \partial}{ \partial r} (r^{2} P_{r}) = 3k r^{2}r \frac{1}{r^2}= 3k[/tex]

    This result seems off, but how can we take the dot product and divergence if we don't know anything about [itex]\vec{r}[/itex]?

    Unless we're just suppose to write down the general divergence theorem, leaving the components as [itex] P_r , P_{\theta}, P_{\phi}[/itex] 584a6e1d449f1af10370a8ffc9a12a28.png

    Where all the angles are constant and pulled out of the integral?
    Last edited: Jul 26, 2013
  2. jcsd
  3. Jul 26, 2013 #2
    Bound and Volume Charges

    You got [itex]\rho _b [/itex] even though you didn't know it. You know a lot of things about [itex]\vec{r}[/itex] in this case, since [itex]\vec{r}=r\hat{r}[/itex]. It is only a function of r, so you can treat [itex]\nabla[/itex] simply as [itex]\frac{ \partial}{ \partial r}[/itex]. Your typing got ugly there, but [itex]\rho _b =-3k [/itex] is correct once you differentiate.

    You were close with [itex]\sigma _b = \vec{P}(\vec{r}) \cdot \hat{n}[/itex] correct. In this case, the normal to the shell is in the radial direction, so [itex]\hat{n} = \hat{r} [/itex] and you can take the dot product very easily, since [itex]\vec{r}=r\hat{r}[/itex]. The bound surface charge is epomenically at the surface, so the magnitude of [itex]\vec{r} = R[/itex], the radius of the sphere.

    So that's the first part of the question. Now you can try to to get part b using the equations you listed.

    Hope that helps. Let me know if I can help more.

    Dr Peter Vaughan
    BASIS Peoria Physics
  4. Jul 27, 2013 #3
    Thanks, it's interesting that I was so close, I wonder if it's luck or intuition :wink:

    So, here it goes.

    ##\vec{P} \cdot \hat{n} = KR##

    $$\frac{KR^3 }{2 \epsilon _0} \int ^{\pi}_{0} \frac{sin \theta '}{\sqrt{R^2+z^2-2RZcos \theta '}} d \theta ' + \frac{3K}{2 \pi \epsilon_0} \int ^{\pi}_0 \int ^{R}_0 \frac{R^2 sin \theta ' }{\sqrt{R^2+z^2-2RZcos \theta '}}dr d \theta ' $$

    Evauliting the first integral gives the term $$\frac{KR^3 }{2 \epsilon _0} ( -\frac{\sqrt{(R-z)+(R+z)}}{Rz})$$, which should equal $$-\frac{KR^3}{z \epsilon _0}$$

    The second term came out wrong, and I'm not exactly sure what happened.


    This is clearly wrong, unless I typed something screwy.
    Last edited: Jul 27, 2013
  5. Jul 27, 2013 #4
    That's messy math.

    I'm not 100% clear on Wolfram Integrate notation, but it looks to my like you are treating R as a variable during integration of r. I believe it is a constant. Both your bound charge and surface charge turned out to be constants. This should make the integration very simple. You might want to check your text and see if there is a simpler form of V that you could be using when [itex]\sigma_b [\itex] and [itex]\rho_b [\itex] are constants. Electrostatics doesn't *usually* involve solving horribly messy integrals unless there's a typo in the problem.
  6. Jul 27, 2013 #5
    Even though ##\sigma _b## ##\rho _b## are constants, the second integral still needs to be integrated over R, because R has to vary inside the sphere (this is a volume integral).

    I admit that I peeked at the solution in the book, and it always seems to skip the integration, and it gives ##\vec{E}## from a different method, other than direct integration.
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