Potential of a polarized hollow spherical shell due to uniform electric field

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:

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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 $$V = – int_{infty}^{r} mathbf E cdot dmathbf z$$ 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?

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

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 $l ne 1?$