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The set up is "A conducting sphere of radius $r_0$ is placed in an originally uniform electric field E, and maintained at zero potential. Show that the potential outside the sphere is:

$$\Phi(r,\theta,\phi)=|E|\bigg(r-\frac{r_0^3}{r^2}\bigg)cos(\theta)$$

where the $\theta=0$ is aligned with the direction of E."

I understand the set up means that E is in the z direction and the boundary conditions are set by the surface of the sphere being at 0 potential, $$\Phi(r_0,\theta,\phi)=0$$ and then I'm a bit stuck. The next condition is at infinity where the sphere does not affect the potential. The potential goes to infinity as r goes to infinity but in what way? Note I have to use these conditions to reduce the general result of $\nabla^2\Phi=0$ in spherical coordinates and not by any other method. This is,

$$\Phi(r,\theta,\phi)=\sum_{l=0}^\infty\sum_{m=0}^l (A_{lm}r^l+B_{lm}r^{-l-1})P_l^m(cos\theta)e^{\pm im \phi}$$

Basically I am a little confused on getting and then applying the second boundary condition. $$r \rightarrow \infty \implies \Phi = -\int_\infty^r \vec{E}\cdot d\vec{l} \rightarrow\infty $$ if the sphere was not present.

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You are not implementing the outer boundary condition correctly. The potential of constant electric field (which I denote by $\Phi_*$) cannot be defined as $\int_ \infty^r E\cdot dr$ because it diverges at infinity. This divergence arises because a constant electric field everywhere in spacetime is not really physical and has infinite energy. However, to cure this problem, we note that the potential is ambiguous up to addition of a constant. In other words it does not matter which reference point we choose for the lower limit of the integral. In this problem, you have to take another reference point at finite distance, e.g. $r=0$. Then the potential is given by $$\Phi_*=-\int_{0}^r E\cdot dr= |E|r \cos \theta$$ if the electric field is in the negative z direction.

Now for the outer boundary condition we require that $$\lim_{r\to\infty}(\Phi(r)-\Phi_*(r))=0$$.
In large $r$, the general multipole expansion you wrote reduces to $$\phi=\sum A_{lm}r^l Y_{lm}$$ Comparing these two equations imply that $$A_{lm}=|E|\,\delta_{l,1}\delta_{m,0},\qquad $$ From the inner boundary condition we conclude $A_{lm}r_0^l+B_{lm}r_0^{-(l+1)}=0$. Therefore $$B_{lm}=-r_0^{2l+1}A_{lm}$$ All these together implies the result you wrote, i.e. $\Phi=|E|(r-r_0^3/r^2)Y_{1,0}$

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