Question

Compare Eqs. 2.15, 4.9, and 6.11. Notice that if $\mathit{\rho }$,$\mathit{P}$ , and $\mathit{M}$are uniform, the same integral is involved in all three:

$\mathbf{\int }\frac{\widehat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}{\mathit{\tau }}^{\mathbf{\text{'}}}$

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 $\mathit{V}$inside and outside a uniformly polarized sphere (Ex. 4.2), and$\mathit{A}$ inside and outside a uniformly magnetized sphere (Ex. 6.1).

Short answer

The inside and outside potential of uniformly magnetized sphere is$\frac{1}{3{\epsilon }_0}Pr$ and$\frac{1}{3{\epsilon }_0}\frac{R^3}{r^2}P\widehat{r}$ respectively. Inside and outside vector potential uniformly magnetized sphere is$\frac{{\mu }_0}{3}(M\times r)$ and $\frac{{\mu }_0}{3}\frac{R^3}{r^2}(M\times \widehat{r})$respectively.

Step-by-step solution

Write the given data from the question.

The charge density is$\rho$ .

The uniform polarization is$P$ .

The integral is $\int \frac{\widehat{r}}{r^2}d{\tau }^{\text{'}}$

The inside the radius of sphere is $r$and outside the radius of the sphere is$R$ .

Determine the general formulas to calculate the V  inside and outside a uniformly polarized sphere and A inside and outside a uniformly magnetized sphere.

The expression for electric field due to uniformly charged sphere is given as follows.

$\mathit{E}\mathbf{=}\mathit{\rho }\left[\frac{1}{4\pi {\epsilon }_0}\int \frac{\widehat{r}}{r^2}d{\tau }^{\text{'}}\right]$ …… (1)

The expression for the potential due to uniformly charged sphere is given as follows.

$\mathit{V}\mathbf{=}\mathit{P}\mathbf{\cdot }\left[\left(\frac{1}{4\pi {\epsilon }_0}\int \frac{\widehat{r}}{r^2}d{\tau }^{\text{'}}\right)\right]$ …… (2)

The expression for the vector potential due to uniformly magnetized sphere is given as follows.

$\mathit{A}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\int \frac{M\times \widehat{r}}{r^2}d{\tau }^{\text{'}}$ …… (3)

Calculate the V inside and outside a uniformly polarized sphere and A inside and outside a uniformly magnetized sphere.

The volume of the sphere is given by,

$\int d{\tau }^{\text{'}}=\frac{4}{3}\pi r^3$

The inside potential due to uniformly charged sphere is given by,

$\begin{array}{l}{V}_{in}=P\cdot \left[\frac{1}{4\pi {\epsilon }_0}\frac{\widehat{r}}{r^2}\frac{4}{3}\pi r^3\right]\\ V_{in}=P\cdot \left(\frac{1}{3{\epsilon }_0}r\right)\\ V_{in}=\frac{1}{3{\epsilon }_0}Pr\end{array}$

The outside potential due to uniformly charged sphere is given by,

$\begin{array}{l}{V}_{out}=P\cdot \left[\frac{1}{4\pi {\epsilon }_0}\frac{\widehat{r}}{r^2}\frac{4}{3}\pi R^3\right]\\ V_{out}=P\cdot \left(\frac{1}{3{\epsilon }_0}\frac{R^3}{r^2}\widehat{r}\right)\\ V_{out}=\frac{1}{3{\epsilon }_0}\frac{R^3}{r^2}P\widehat{r}\end{array}$

The vector potential due to uniformly magnetized sphere inside the sphere is given by

$\begin{array}{l}{A}_{in}=\frac{{\mu }_0}{4\pi }\int \frac{M\times \widehat{r}}{r^2}d{\tau }^{\text{'}}\\ A_{in}=\frac{{\mu }_0}{4\pi }\frac{M\times \widehat{r}}{r^2}\times \frac{4}{3}\pi r^3\\ A_{in}=\frac{{\mu }_0}{3}(M\times r)\end{array}$

The vector potential due to uniformly magnetized sphere outside the sphere is given by

$\begin{array}{l}{A}_{out}=\frac{{\mu }_0}{4\pi }\int \frac{M\times \widehat{r}}{r^2}d{\tau }^{\text{'}}\\ A_{out}=\frac{{\mu }_0}{4\pi }\frac{M\times \widehat{r}}{r^2}\times \frac{4}{3}\pi R^3\\ A_{out}=\frac{{\mu }_0}{3}\frac{R^3}{r^2}(M\times \widehat{r})\end{array}$

Hence the inside and outside potential of uniformly magnetized sphere is$\frac{1}{3{\epsilon }_0}Pr$ and $\frac{1}{3{\epsilon }_0}\frac{R^3}{r^2}P\widehat{r}$ respectively. Inside and outside vector potential uniformly magnetized sphere is $\frac{{\mu }_0}{3}(M\times r)$and $\frac{{\mu }_0}{3}\frac{R^3}{r^2}(M\times \widehat{r})$ respectively.