Question 6.26: Compare Eqs. 2.15, 4.9, and 6.11. Notice that if ρ, P, and M...

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).

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Eq. 2.15 :  E =\rho\left\{\frac{1}{4 \pi \epsilon_{0}} \int_{ \nu } \frac{\hat{ ᴫ }}{ᴫ^{2}} d \tau^{\prime}\right\}      (for uniform charge density);

Eq. 4.9 :  V= P \cdot\left\{\frac{1}{4 \pi \epsilon_{0}} \int_{ \nu } \frac{\hat{ ᴫ}}{ᴫ^{2}} d \tau^{\prime}\right\}     (for uniform polarization);

Eq. 6.11 :  A =\mu_{0} \epsilon_{0} M \times\left\{\frac{1}{4 \pi \epsilon_{0}} \int_{ \nu } \frac{\hat{ ᴫ }}{ᴫ^{2}} d \tau^{\prime}\right\}           (for uniform magnetization).

For a uniformly charged sphere (radius R): \begin{cases} E _{\text {in }}=\rho\left(\frac{1}{3 \epsilon_{0}} r \right) & \text { (Prob. 2.12) } \\ E _{\text {out }}=\rho\left(\frac{1}{3 \epsilon_{0}} \frac{R^{3}}{r^{2}} \hat{ r }\right) & (\text { Ex. 2.3 })\end{cases}   (Prob. 2.12)   (Ex. 2.3)

So the scalar potential of a uniformly polarized sphere is: \left\{\begin{array}{l}V_{ in }=\frac{1}{3 \epsilon_{0}}( P \cdot r ) ,\\V_{ out }=\frac{1}{3 \epsilon_{0}} \frac{R^{3}}{r^{2}}( P \cdot \hat{ r }), \end{array}\right.

and the vector potential of a uniformly magnetized sphere is: \left\{\begin{array}{l} A _{\text {in }}=\frac{\mu_{0}}{3}( M \times r ), \\A _{\text {out }}=\frac{\mu_{0}}{3} \frac{R^{3}}{r^{2}}( M \times \hat{ r }),\end{array}\right.

(confirming the results of Ex. 4.2 and of Exs. 6.1 and 5.11).

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