EM M97 1

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A spherical shell is uniformly polarized. The electric polarization per unit volume is \mathbf{P}=P\hat{\mathbf{k}}, and the inner and outer radii of the shell are a and b. Find the electric potential in each of the three regions: r>b, a<r<b, and r<a.

The effective charge density is given by:

\rho_{eff}=\rho-\mathbf{\nabla}\cdot\mathbf{P}

Where ρ is the free space charge density and P is the polarization. Since there is no free space charge density, and the polarization is constant except on boundaries, we can write the effective surface charge density on the spheres as:

\sigma_{pol}=-\left(\mathbf{P}_{in}-\mathbf{P}_{out}\right)\cdot\hat{\mathbf{r}}

Where \mathbf{P}_{out} is the polarization on the outside sphere, and \hat{\mathbf{r}} is the unit vector in the radial direction. So:

\sigma_{r=a}=-\left(-P\hat{\mathbf{k}}\right)\bullet\hat{\mathbf{r}}=P\cos\theta
\sigma_{r=b}=-\left(P\hat{\mathbf{k}}\right)\bullet\hat{\mathbf{r}}=-P\cos\theta

Where we have chosen the θ = 0 direction to be in the \hat{\mathbf{k}} direction, so that θ is the angle between \hat{\mathbf{k}}. The general solution for the Laplace equation in spherical coordinates with azimuthal symmetry is:

\Phi(r,\theta)=\sum_{l=0}^{\infty}\left[A_{l}r^{l}+B_{l}r^{-(l+1)}\right]P_{l}(\cos\theta)

Where Pl is the l'th Legendre polynomial. Obviously our solution should use l=1:

\Phi(r,\theta)=\left[Ar+Br^{-2}\right]\cos\theta

So it is just a matter of matching the boundary conditions. If we label the regions I (r < a), II (a < r < b) and III (r > b), it will make things easier. Since we must have reasonable behavior as r\rightarrow\infty :

\Phi_{I}(r,\theta)=A\left(\frac{r}{a}\right)\cos\theta
\Phi_{II}(r,\theta)=\left[B\left(\frac{r}{a}\right)+C\left(\frac{b}{r}\right)^{2}\right]\cos\theta
\Phi_{III}(r,\theta)=D\left(\frac{b}{r}\right)^{2}\cos\theta

We now find boundary conditions at r=a . Continuity:

\left.\Phi_{I}\right|_{r=a}=\left.\Phi_{II}\right|_{r=a}
A\cos\theta=\left[B+C\left(\frac{b}{a}\right)^{2}\right]\cos\theta
A=B+C\left(\frac{b}{a}\right)^{2}

And derivative:

\epsilon_{0}\left.\frac{\partial\Phi_{I}}{\partial r}\right|_{r=a}-\epsilon_{0}\left.\frac{\partial\Phi_{II}}{\partial r}\right|_{r=a}=P\cos\theta
\frac{A}{a}\cos\theta-\left[\frac{B}{a}-\frac{2C}{a}\left(\frac{b}{a}\right)^{2}\right]\cos\theta=\frac{P}{\epsilon_{0}}\cos\theta
A-B+2C\left(\frac{b}{a}\right)^{2}=\frac{Pa}{\epsilon_{0}}

And for r=b , Continuity:

\left.\Phi_{II}\right|_{r=b}=\left.\Phi_{III}\right|_{r=b}
\left[B\frac{b}{a}+C\right]\cos\theta=D\cos\theta
B\frac{b}{a}+C=D

And derivative:

\epsilon_{0}\left.\frac{\partial\Phi_{II}}{\partial r}\right|_{r=a}-\epsilon_{0}\left.\frac{\partial\Phi_{III}}{\partial r}\right|_{r=a}=-P\cos\theta
\left[\frac{B}{a}-\frac{2C}{b}\right]\cos\theta+\frac{2D}{b}\cos\theta=-\frac{P}{\epsilon_{0}}\cos\theta
B\frac{b}{a}-2C+2D=-\frac{Pb}{\epsilon_{0}}

Collecting all of our equations:

\begin{matrix}A &amp; = &amp; B+C\left(\frac{b}{a}\right)^{2} \\ A-B+2C\left(\frac{b}{a}\right)^{2} &amp; = &amp; Pa/\epsilon_{0} \\ B\frac{b}{a}+C &amp; = &amp; D \\ B\frac{b}{a}-2C+2D &amp; = &amp; -Pb/\epsilon_{0}\end{matrix}

Solving the first two together for C :

3C\left(\frac{b}{a}\right)^{2}=Pa/\epsilon_{0}\Rightarrow C=\frac{Pa}{3\epsilon_{0}}\left(\frac{a}{b}\right)^{2}

Solving the last two for B :

3B\frac{b}{a}=-Pb/\epsilon_{0}\Rightarrow B=-\frac{Pa}{3\epsilon_{0}}

And plugging onto the continuity equations:

A=-\frac{Pa}{3\epsilon_{0}}+\frac{Pa}{3\epsilon_{0}}\left(\frac{a}{b}\right)^{2}\left(\frac{b}{a}\right)^{2}=0
D=-\frac{Pa}{3\epsilon_{0}}\frac{b}{a}+\frac{Pa}{3\epsilon_{0}}\left(\frac{a}{b}\right)^{2}=\frac{P}{3\epsilon_{0}b^{2}}\left(a^{3}-b^{3}\right)

Plugging in:

\begin{matrix}\Phi(r&lt;a) &amp; = &amp; 0 \\  \Phi(a&lt;r&lt;b) &amp; = &amp; \left[-\frac{P}{3\epsilon_{0}}r+\frac{Pa}{3\epsilon_{0}}\left(\frac{a}{r}\right)^{2}\right]\cos\theta \\ \Phi(r&gt;b) &amp; = &amp; \frac{P}{3\epsilon_{0}}\left(a^{3}-b^{3}\right)r^{-2}\cos\theta \end{matrix}

This page was recovered in October 2009 from the Plasmagicians page on Prelim_M97_EM1 dated 22:52, 18 December 2005.

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