A hollow spherical shell has inner radius $a$, outer radius $b$ and is made with a material with dielectric constant $\epsilon_r = \epsilon / \epsilon_0$. It is placed in a uniform electric field $\mathbf E(\mathbf r) = E_0 \hat z$ and becomes polarized.

I am meant to determine the electric potential $V$ at all points in space, with the assistance of an attached solution to it.

My thought process is this - please let me know if anything I've stated here is not logically sound:

Firstly, since the electric field is uniform, the polarization will thus be uniform. This means there will be no volume bound charge density - and just a surface bound charge density, which ought to be located at $r = b$. Since there is no discussion of charge placed anywhere, there is no free charge. Thus since $\rho_b = \rho_f = 0$, I will state that $\rho = 0$. Thus, there is no charge anywhere except $r=b$ and namely no charges at $r<a, a<r<b,$ and $r>b$. Those regions are thus solutions to Laplace's equation.

With this in mind, I will need to establish some reasonable boundary conditions. I say these are reasonable:

$$V(r=0, \theta) \ \text{is finite}$$

As this doesn't sound like it makes physical sense if untrue. Additionally,

$$V(r >> b, \theta) \ \text{is the potential due to the uniform $\mathbf E$ field}$$

I don't know how to find the potential due to this uniform field by computing $$V = - \int_{\infty}^{r} \mathbf E \cdot d\mathbf z$$ since this diverges, however just looking at

$$-\nabla V = E_0 \hat z$$

Tells me that $$\implies V = - E_0 z = -E_0\ r \cos{\theta}$$

But I can't justify it otherwise if it wasn't a basic field.

Apart from that, I begin with the general solution to Laplace's equation in spherical coordinates (noting there is azimuthal symmetry):

$$V(r,\theta) = \sum_{l=0}^\infty \left(A_l r^l + \frac{B_l}{r^{l+1}}\right) P_l (\cos{\theta})$$

I then try and use my boundary conditions to make my answer for $V$ in each region a bit more specific.

For $r < a$, I say $B_l = 0$ to prevent this diverging at $r=0$ consistent with my first boundary condition, so I have

$$ V_1(r<a,\theta) = \sum_{l=0}^\infty A_l r^l P_l (\cos{\theta}) $$

For $a<r<b$ I can't make any assumptions given the boundary conditions, so I have

$$V_2(a<r<b,\theta) = \sum_{l=0}^\infty \left(B_l r^l + \frac{C_l}{r^{l+1}}\right) P_l (\cos{\theta})$$

For $r>b$ I impose the second boundary condition. Therefore I need my solution to converge to $V(r>>b,\theta) \to -E_0 r \cos{\theta}$ which means I need to take out the $A_l$ term in my Laplace general solution.

$$V_3(r>b,\theta) = -E_0 r \cos{\theta} + \sum_{l=0}^\infty \left(\frac{E_l}{r^{l+1}}\right) P_l (\cos{\theta})$$

Next, I note that potential suffers no discontinuities in electromagnetism, as it's just a sum of scalars. This means I can state..

$$V_1(a,\theta) = V_2(a,\theta)$$ $$V_2(b,\theta) = V_3(b, \theta)$$

$$\implies \sum_{l=0}^\infty A_l a^l P_l (\cos{\theta}) = \sum_{l=0}^\infty \left(B_l a^l + \frac{C_l}{a^{l+1}}\right) P_l (\cos{\theta})$$

$$\implies \sum_{l=0}^\infty \left(B_l b^l + \frac{C_l}{b^{l+1}}\right) P_l (\cos{\theta}) = -E_0 b \cos{\theta} + \sum_{l=0}^\infty \left(\frac{E_l}{b^{l+1}}\right) P_l (\cos{\theta})$$

From here, I feel I am liberty to then say, $\forall \ l$:

$$A_l a^l = B_l a^l + \frac{C_l}{a^{l+1}}$$

However, my lecturer writes in the solution:

For $l=1$:

$$B_1 b + C_1/b^2 = -E_0b + D_1/b^2 \ \ \text{because $\cos{\theta} = P_1 (\cos{\theta})$}$$

For $l \ne 1$:

$$B_l b + C_l/b^{l+1} = -E_0b + D_l/b^{l+1}$$

However, I don't see how $$\sum_{l=0}^\infty \left(B_l b^l + \frac{C_l}{b^{l+1}}\right) P_l (\cos{\theta}) = -E_0 b \cos{\theta} + \sum_{l=0}^\infty \left(\frac{E_l}{b^{l+1}}\right) P_l (\cos{\theta})$$

means you can extract

$$B_l b + C_l/b^{l+1} = -E_0b + D_l/b^{l+1}$$

like the way it was done in

$$\sum_{l=0}^\infty A_l a^l P_l (\cos{\theta}) = \sum_{l=0}^\infty \left(B_l a^l + \frac{C_l}{a^{l+1}}\right) P_l (\cos{\theta})$$

Because of the stray $-E_0 r \cos{\theta}$ term. Secondly, where did the cosine term go? I thought, if anything,

$$\sum_{l=0}^\infty \left(B_l b^l + \frac{C_l}{b^{l+1}}\right) P_l (\cos{\theta}) = -E_0 b \cos{\theta} + \sum_{l=0}^\infty \left(\frac{E_l}{b^{l+1}}\right) P_l (\cos{\theta}) $$

$$\implies B_l b + C_l/b^{l+1} = -E_0b \cos{\theta} + D_l/b^{l+1}$$

Finally, my lecturer notes:

For now, let's just work out the $l=1$ coefficients. The $l \ne 1$ coefficients will have the same form for $l=1$ except we would use $E_0 = 0$. We'll show that all $l=1$ coefficents are proportional to $E_0$, which shows that $l \ne 1$ coefficients are zero.

Why is $E_0$ if $l \ne 1$? Even if the $l=1$ coefficients are proportional to $E_0$, why does this imply the previous sentence? Is it because of Legendre polynomial orthogonality?

Ultimately, the purpose of this is to show the potential has the following form:

enter image description here

And my confusions in doing so are the following:

1) Whether my thought process has holes in it and where

2) Since there is a surface charge density, can I assume it's zero? Can I assume this shell is uncharged?

3) Why does the cosine term vanish where it did and why can I state $B_l b + C_l/b^{l+1} = -E_0b \cos{\theta} + D_l/b^{l+1}$?

4) Why does $E_0 = 0$ if $l \ne 1?$


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