91Ó°ÊÓ

There is a configuration of charges and currents confined within a volume Vand total dipole moment $\stackrel{â‡\P Ä}{p}$.

For a charge density p spread over a volume V, the dipole moment

$\stackrel{\mathbf{â‡\P Ä}}{\mathbf{p}}\mathbf{=}{\mathbf{∫}}_{\mathbf{v}}\mathbf{p}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{r}}\mathbf{d}\mathbf{Ï„}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, r is a spherical polar coordinate and $\mathrm{Ï„}$is an infinitesimal volume element.

From equation (1), the rate of change of dipole moment

$\frac{d\stackrel{â‡\P Ä}{p}}{dt}=\frac{d}{dt}{∫}_v\mathrm{ÒÏ}\stackrel{â‡\P Ä}{r}d\mathrm{Ï„}$

Use continuity equation to get

$\frac{\mathrm{d}\stackrel{â‡\P Ä}{\mathrm{p}}}{\mathrm{dt}}=-{∫}_{\mathrm{v}}\left(\stackrel{â‡\P Ä}{∇}.\stackrel{â‡\P Ä}{\mathrm{J}}\right)\stackrel{â‡\P Ä}{\mathrm{r}}\mathrm{»åÏ„}$

Here, $\stackrel{â‡\P Ä}{J}$is the current density.

Use just one component of $\stackrel{â‡\P Ä}{r}$and product rule to prove

$$\begin{gathered} {∫}_{\mathrm{v}}\left(\stackrel{â‡\P Ä}{∇}.\stackrel{â‡\P Ä}{\mathrm{J}}\right)x\mathrm{»åÏ„}=-{∫}_{\mathrm{v}}\stackrel{â‡\P Ä}{∇}.\left(x\stackrel{â‡\P Ä}{\mathrm{J}}\right)d\mathrm{Ï„}+{∫}_{\mathrm{v}}\left(\stackrel{â‡\P Ä}{∇}x\right).\stackrel{â‡\P Ä}{\mathrm{J}}d\mathrm{Ï„} \\ =-{âˆ\circledR }_{s}x\stackrel{â‡\P Ä}{\mathrm{J}}.d\stackrel{â‡\P Ä}{s}+{∫}_v\hat{x}.\stackrel{â‡\P Ä}{\mathrm{J}}d\mathrm{Ï„} \\ ={∫}_v{J}_{x}d\mathrm{Ï„} \end{gathered}$$

Here, the surface integral goes to zero because the surface can be taken at infinity where the current density is zero. The x component of the current density is left inside the integral.

Considering the other two components,

$\frac{d\stackrel{â‡\P Ä}{p}}{dt}={∫}_v\stackrel{â‡\P Ä}{J}d\mathrm{Ï„}$

Hence, the equation is proved.