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Astrum
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Homework Statement
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
Homework 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]
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]
Where all the angles are constant and pulled out of the integral?
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