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Potential of a uniformly polarized sphere

  1. Mar 25, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the potential of a uniformly polarized sphere directly from eq. 9

    2. Relevant equations

    [tex]V(r)=k \int \frac {P(r') \cdot \hat{r}} {r^2} d\tau[/tex]
    3. The attempt at a solution
    P is a constant and can be factored out. Since r is taken, call the radius of the sphere R and and an arbitrary radial length l. Then by the law of cosines we can express the denominator for every volume element as.

    [tex]r^2=l^2+z^2-2lzcos\theta[/tex]

    This is where I'm stuck. What can I do with r hat? Nothing? I understand that its a unit vector in the direction of of the volume element to the point I'm trying to evaluate at, but I'm not sure if I need to change anything about it or not. z is constant so in theory once I have that changed I can evaluate the integral. Is there an easier way?
     
  2. jcsd
  3. Mar 25, 2017 #2
    It says uniformly polarized, but what is the direction of polarization? Keep in mind it's actually a vector field, and you are dotting it with ##\hat r##.
     
  4. Mar 25, 2017 #3

    kuruman

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    Your expression for the potential should be
    $$ V(r)=k \int \frac {\vec{P}(r') \cdot (\vec{r}-\vec{r}')} {|\vec{r}-\vec{r}'|^{3/2}} d^3 r' $$
    where ##\vec{r}## is the position of the point of interest and ##\vec{r}'## is the position of volume element ##d^3 r'##. Note the integration over primed coordinates. I suggest that you write each of the vectors in the integrand in terms of unit vectors, then assemble the integrand.
     
  5. Mar 26, 2017 #4
    Okay so P is uniform in the z direction. z hat dotted with r hat is equal to the cosine of the angle between the two, which I call phi. This angle is the top angle in the triangle that i utilized with the law of cosines to find r^2. Using the law of cosines again, I can relate phi to the rest of the triangle and plug in for r.

    [tex]
    l^{2}=z^2+r^2-2zrcos\phi[/tex]

    [tex]cos\phi = \frac {l^2-z^2-r^2} {-2zr}[/tex]

    With some simplication,

    [tex]kP \int \frac {z - lcos\theta} {(z^2+l^2-2zlcos\theta)^{3/2}}l^2sin\theta dld\theta d\psi[/tex]

    How's that integral look?
     
  6. Mar 26, 2017 #5

    kuruman

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    You are cutting corners that will lead you into trouble. The integration is over primed coordinates which your integrand does not show.
    1. Use spherical coordinates ##(r, \theta, \phi)## for ##\vec{r}## and ##(r', \theta ', \phi ')## for ##\vec{r}'##.
    2. Write expressions for ##\vec{r}## and ##\vec{r}'## in terms of these coordinates in the Cartesian representation. Without loss of generality, you can set ##\phi=0## and ##\phi ' = 0##.
    3. Find an expression for ##(\vec{r}-\vec{r}') ##.
    4. Find an expression for ##|\vec{r}-\vec{r}'|^{3/2} ##.
    5. Take the dot product ##\vec{P}(r') \cdot (\vec{r}-\vec{r}')##.
    6. Assemble the integrand. It should be in terms of ##(r, \theta, \phi)## and ##(r', \theta ', \phi ')##.
    7. Integrate over primed coordinates. I would do ##\theta '## first.
     
  7. Mar 26, 2017 #6
    kuruman is right. I didn't notice at first but your equation is slightly off. But, if that's what the book is giving you/asking you to use then so be it.
     
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