91Ó°ÊÓ

Consider the given defining equations for $A(viz.∇.A=0,∇×A=B)$.

Consider the given Maxwell's equations for$B(viz.∇.B=0,∇×B={\mathrm{μ}}_{0}J)$ .

Write the formula of vector potential.

$\mathbf{∇}\mathbf{×}\mathit{B}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{J}$ …… (1)

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

Write the formula of exploiting the appropriate given analogy.

$\mathit{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{P}\mathbf{(}\mathbf{r}\mathbf{\text{'}}\mathbf{)}\mathbf{-}\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathit{d}\mathit{Ï„}\mathbf{\text{'}}$ …… (2)

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

Let the magnetic field strength is B.

We know that

$$\begin{gathered} ∇×B={\mathrm{μ}}_{0}J \\ ∇.B=0 \end{gathered}$$

In the integral form the magnetic field is given by,

$B=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\frac{J×\hat{r}}{r^2}d\mathrm{τ}\text{'}$

The expressions for vector potential is given as

localid="1658557591266" $∇×A=B$

And

$∇.A=0$

Determine the vector potential.

Substitute$∇×A$for B into equation (1).

$∇×\left(∇×A\right)={\mathrm{μ}}_{0}J$

Use vector identity,

${∇}^{2}A=∇\left(∇.A\right)-∇×\left(∇×A\right)$

Substitute 0 for$∇.A$and${\mathrm{μ}}_{0}J$for $∇×\left(∇×A\right)$

${∇}^{2}A=-{\mathrm{μ}}_{0}J$

Therefore, the vector potential can be written as$A=\frac{1}{4\mathrm{π}}∫\frac{B×\hat{r}}{r^2}d\mathrm{τ}\text{'}$.

Poisson equation is given by,

${∇}^{2}V=-\frac{-P}{{\mathrm{Î\mu }}_0}$

Volume charge density is given by,

$p_b=-∇.P$

Compare both charge densities so,

$P=-{\mathrm{Î\mu }}_{0}E$

Determine the potential is given by,

Substitute$-{\mathrm{Î\mu }}_0$ for P into equation (2).

$V\left(r\right)=-\frac{1}{4\mathrm{π}}∫\frac{E\left(r\text{'}\right).\hat{r}}{r^2}d\mathrm{τ}\text{'}$

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