(a) one way to fill in the "missing link" in Fig. 5.48 is to exploit the analogy between the defining equations for $\mathit{A}\mathbf{(}\mathit{v}\mathit{i}\mathit{z}\mathbf{\nabla }\mathbf{-}\mathit{A}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\nabla }\mathbf{\times }\mathit{A}\mathbf{=}\mathit{B}\mathbf{)}$and Maxwell's equations for$\mathit{B}\mathbf{(}\mathit{v}\mathit{i}\mathit{z}\mathbf{\nabla }\mathbf{.}\mathit{B}\mathbf{=}\mathbf{0}\mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathbf{}\mathit{J}\mathbf{)}$.Evidently A depends on B in exactly the same way that B depends on${\mathit{\mu }}_{\mathbf{0}}\mathbf{}\mathit{J}$(to wit: the Biot-Savart law). Use this observation to write down the formula for A in terms of B.

(b) The electrical analog to your result in (a) is

localid="1658557463395" $\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }}\mathbf{\int }\frac{\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{\text{'}}\mathbf{)}\mathbf{-}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathit{\tau }\mathbf{\text{'}}$

Derive it, by exploiting the appropriate analogy.

Consider the given defining equations for $A(viz.\nabla .A=0,\nabla \times A=B)$.

Consider the given Maxwell's equations for$B(viz.\nabla .B=0,\nabla \times B={\mu }_{0}J)$ .

Write the formula of vector potential.

$\mathbf{\nabla }\mathbf{\times }\mathit{B}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}\mathit{J}$ …… (1)

Here,${\mathit{\mu }}_{\mathbf{0}}$is permeability and$\mathit{J}$is current density.

Write the formula of exploiting the appropriate given analogy.

$\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{P}\mathbf{(}\mathbf{r}\mathbf{\text{'}}\mathbf{)}\mathbf{-}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathit{\tau }\mathbf{\text{'}}$ …… (2)

Here,$\mathit{P}$is volume charge density, $\hat{\mathbf{r}}$is radius of spherical shell,${\mathit{\epsilon }}_{\mathbf{0}}$is relative pemitivity.

Let the magnetic field strength is B.

We know that

$$\begin{gathered} \nabla \times B={\mu }_{0}J \\ \nabla .B=0 \end{gathered}$$

In the integral form the magnetic field is given by,

$B=\frac{{\mu }_0}{4\pi }\int \frac{J\times \hat{r}}{r^2}d\tau \text{'}$

The expressions for vector potential is given as

localid="1658557591266" $\nabla \times A=B$

And

$\nabla .A=0$

Determine the vector potential.

Substitute$\nabla \times A$for B into equation (1).

$\nabla \times \left(\nabla \times A\right)={\mu }_{0}J$

Use vector identity,

${\nabla }^{2}A=\nabla \left(\nabla .A\right)-\nabla \times \left(\nabla \times A\right)$

Substitute 0 for$\nabla .A$and${\mu }_{0}J$for $\nabla \times \left(\nabla \times A\right)$

${\nabla }^{2}A=-{\mu }_{0}J$

Therefore, the vector potential can be written as$A=\frac{1}{4\pi }\int \frac{B\times \hat{r}}{r^2}d\tau \text{'}$.

Poisson equation is given by,

${\nabla }^{2}V=-\frac{-P}{{\epsilon }_0}$

Volume charge density is given by,

$p_b=-\nabla .P$

Compare both charge densities so,

$P=-{\epsilon }_{0}E$

Determine the potential is given by,

Substitute$-{\epsilon }_0$ for P into equation (2).

$V\left(r\right)=-\frac{1}{4\pi }\int \frac{E\left(r\text{'}\right).\hat{r}}{r^2}d\tau \text{'}$

Therefore, the value of exploiting the appropriate analogy is $V\left(r\right)=-\frac{1}{4\pi }\int \frac{E\left(r\text{'}\right).\hat{r}}{r^2}$.