The general solution for the electric potential in spherical coordinates with azimuthal symmetry (no $\phi$ dependence) is: $$V(r,\theta) = \sum_{l=0}^{\infty} \left(A_l r^l + \dfrac{B_l}{r^{l+1}}\right)P_l(\cos \theta)$$ Consider a metal sphere (constant potential in and on the sphere, remember). Which terms in the sum vanish outside the sphere? (Recall: $V \rightarrow 0$ as $r \rightarrow \infty$) 1. All the $A_l$'s 2. All the $A_l$'s except $A_0$ 3. All the $B_l$'s 4. All the $B_l$'s except $B_0$ 5. Something else Note: * CORRECT ANSWER: E * Only B0 will survive.

Given $V_0(\theta) = \sum_l C_l P_l(\cos \theta)$, we want to get to the integral: $$ \int_{-1}^{+1}P_l(u)\;P_m(u)\;du = \dfrac{2}{2+1}\; (\mathrm{for}\;l=m)$$ we can do this by multiplying both sides by: 1. $P_m(\cos \theta)$ 2. $P_m(\sin \theta)$ 3. $P_m(\cos \theta) \sin \theta$ 4. $P_m(\sin \theta) \cos \theta$ 5. $P_m(\sin \theta) \sin \theta$ Note: * CORRECT ANSWER: D

$$V(r,\theta) = \sum_{l=0}^{\infty} \left(A_l r^l + \dfrac{B_l}{r^{l+1}}\right)P_l(\cos \theta)$$ Suppose V on a spherical shell is: $$V(R,\theta) = V_0 \left(1+\cos^2\theta\right)$$ Which terms do you expect to appear when finding **V(inside)**? 1. Many $A_l$ terms (but no $B_l$'s) 2. Many $B_l$ terms (but no $A_l$'s) 3. Just $A_0$ and $A_2$ 4. Just $B_0$ and $B_2$ 5. Something else! Note: * CORRECT ANSWER: C

$$V(r,\theta) = \sum_{l=0}^{\infty} \left(A_l r^l + \dfrac{B_l}{r^{l+1}}\right)P_l(\cos \theta)$$ Suppose V on a spherical shell is: $$V(R,\theta) = V_0 \left(1+\cos^2\theta\right)$$ Which terms do you expect to appear when finding **V(outside)**? 1. Many $A_l$ terms (but no $B_l$'s) 2. Many $B_l$ terms (but no $A_l$'s) 3. Just $A_0$ and $A_2$ 4. Just $B_0$ and $B_2$ 5. Something else! Note: * CORRECT ANSWER: D

How many boundary conditions (on the potential $V$) do you use to find $V$ inside the spherical plastic shell? <img src="./images/plastic_shell_vtheta.png" align="right" style="width: 350px";/> 1. 1 2. 2 3. 3 4. 4 5. It depends on $V_0(\theta)$ Note: * CORRECT ANSWER: B * Good for discussion; obviously you need the surface BC, but what about at r=0? Is that technically a BC?