Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ, P, and M are uniform, the same integral is involved in all three:
E ( r )=\frac{1}{4 \pi \epsilon_{0}} \int_{\text {all space }} \frac{\hat{ᴫ}}{ᴫ^{2}} \rho\left( r ^{\prime}\right) d \tau^{\prime} (2.15)
V( r )=\frac{1}{4 \pi \epsilon_{0}} \int_{ \nu } \frac{ P \left( r ^{\prime}\right) \cdot \hat{ᴫ}}{ᴫ^{2}} d \tau^{\prime} (4.9)
A ( r )=\frac{\mu_{0}}{4 \pi} \int \frac{ M \left( r ^{\prime}\right) \times \hat{ᴫ}}{ᴫ^{2}} d \tau^{\prime} (6.11)
\int \frac{\hat{ᴫ}}{ᴫ^{2}} d \tau^{\prime} .
Therefore, if you happen to know the electric field of a uniformly charged object, you can immediately write down the scalar potential of a uniformly polarized object, and the vector potential of a uniformly magnetized object, of the same shape. Use this observation to obtain V inside and outside a uniformly polarized sphere (Ex. 4.2), and A inside and outside a uniformly magnetized sphere (Ex. 6.1).