# 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:

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 $$r and $$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

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

$$V_2(a

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:

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 $$l \ne 1?$$