91Ó°ÊÓ

The charge density is$\mathrm{ÒÏ}\left(r,t\right)$ .

The current density is$j\left(r\right)$ .

The expression for the continuity equation is given as follows.

$\frac{\mathbf{∂}\mathbf{ÒÏ}}{\mathbf{∂}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{∇}\mathbf{×}\mathit{J}$

The expression for Ampere’s law with Maxwell’s displacement current is given as follows.

$\mathbf{∇}\mathbf{×}\mathit{B}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{J}\mathbf{+}{\mathit{μ}}_{\mathbf{0}}{\mathit{Î\mu }}_{\mathbf{0}}\frac{\mathbf{∂}\mathbf{E}}{\mathbf{∂}\mathbf{t}}$

Here, B is the magnetic field, E is the electric field, ${\mathrm{μ}}_0$is the free space permeability and is the permittivity.

(a)

Consider the expression for the continuity equation.

$\mathrm{ÒÏ}\left(t\right)=\left(-∇·J\right)t+\text{constant}$ ……. (1)

Here the constant should be a function of but not the .

Therefore, $\mathrm{ÒÏ}\left(r,0\right)$

Substitute $\mathrm{ÒÏ}\left(r,0\right)$ for constant$\mathrm{ÒÏ}\left(r,0\right)$ and for data-custom-editor="chemistry" $-∇·J$into equation (1).

$\mathrm{ÒÏ}\left(\mathrm{t}\right)=\mathrm{ÒÏ}(\mathrm{r},0)t+\mathrm{ÒÏ}\left(t\right)$

Hence the required equation is proved.

(b)

The equation of ampere’s law with the Maxwell’s displacement term is given by,

$$\begin{gathered} ∇×B={\mathrm{μ}}_{0}J+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t} \\ ∇×B={\mathrm{μ}}_{0}J-\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}∫J·∇\frac{\hat{r}}{r^2}d\mathrm{Ï„} \end{gathered}$$ …… (2)

Consider the condition,

$-\left(J·∇\right)\frac{\hat{r}}{r^2}=\left(J·∇\right)\frac{\hat{r}}{r^2}d\mathrm{τ}$

Substitute$-\left(J·∇\right)\frac{\hat{r}}{r^2}$for $-\left(J·∇\right)\frac{\hat{r}}{r^2}$into equation (2).

$$\begin{gathered} ∇×B={\mathrm{μ}}_{0}J+\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\left(J·∇\text{'}\right)\frac{\hat{r}}{r^2}d\mathrm{τ} \\ ∇×B={\mathrm{μ}}_{0}J-\frac{{\mathrm{μ}}_0}{4\mathrm{π}}∫\left(J·∇\text{'}\right)\frac{\hat{r}}{r^2}d\mathrm{τ} \end{gathered}$$

Multiply and divide into the second terms of the right side in the above equation.

$$\begin{gathered} ∇×B={\mathrm{μ}}_{0}J+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{∂}{∂t}∫\frac{\mathrm{ÒÏ}\hat{r}}{r^2}d\mathrm{Ï„} \\ ∇×B={\mathrm{μ}}_{0}J+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂}{∂t}∫\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{ÒÏ}\hat{r}}{r^2}d\mathrm{Ï„} \end{gathered}$$

The term $∫\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\mathrm{ÒÏ}\hat{r}}{r^2}d\mathrm{Ï„}$represent the electric filed into above equation.

Therefore,

$∇×B={\mathrm{μ}}_{0}J+{\mathrm{μ}}_0{\mathrm{Î\mu }}_0\frac{∂E}{∂t}$

Hence, the equation for Ampere’s law with Maxwell’s displacement current term is obtained.