Question

For a configuration of charges and currents confined within a volume

V,show that

${\mathbf{\int }}_{\mathbf{V}}^{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{}}}\mathbf{Jd\tau }\mathbf{=}\frac{\mathbf{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}}{\mathbf{dt}}$

where $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}$is the total dipole moment.

Short answer

It is proved that${\int }_{\mathrm{V}}^{\stackrel{\rightharpoonup }{}}\mathrm{Jd\tau }=\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{p}}}{\mathrm{dt}}$.

Step-by-step solution

Given data

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

Dipole moment

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

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{p}}\mathbf{=}{\mathbf{\int }}_{\mathbf{v}}\mathbf{p}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{d}\mathbf{\tau }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, r is a spherical polar coordinate and $\tau$is an infinitesimal volume element.

Rate of change of dipole moment

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

$\frac{d\stackrel{\rightharpoonup }{p}}{dt}=\frac{d}{dt}{\int }_v\rho \stackrel{\rightharpoonup }{r}d\tau$

Use continuity equation to get

$\frac{\mathrm{d}\stackrel{\rightharpoonup }{\mathrm{p}}}{\mathrm{dt}}=-{\int }_{\mathrm{v}}\left(\stackrel{\rightharpoonup }{\nabla }.\stackrel{\rightharpoonup }{\mathrm{J}}\right)\stackrel{\rightharpoonup }{\mathrm{r}}\mathrm{d\tau }$

Here, $\stackrel{\rightharpoonup }{J}$is the current density.

Use just one component of $\stackrel{\rightharpoonup }{r}$and product rule to prove

$$\begin{gathered} {\int }_{\mathrm{v}}\left(\stackrel{\rightharpoonup }{\nabla }.\stackrel{\rightharpoonup }{\mathrm{J}}\right)x\mathrm{d\tau }=-{\int }_{\mathrm{v}}\stackrel{\rightharpoonup }{\nabla }.\left(x\stackrel{\rightharpoonup }{\mathrm{J}}\right)d\tau +{\int }_{\mathrm{v}}\left(\stackrel{\rightharpoonup }{\nabla }x\right).\stackrel{\rightharpoonup }{\mathrm{J}}d\tau \\ =-{\oint }_{s}x\stackrel{\rightharpoonup }{\mathrm{J}}.d\stackrel{\rightharpoonup }{s}+{\int }_v\hat{x}.\stackrel{\rightharpoonup }{\mathrm{J}}d\tau \\ ={\int }_v{J}_{x}d\tau \end{gathered}$$

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,

$\frac{d\stackrel{\rightharpoonup }{p}}{dt}={\int }_v\stackrel{\rightharpoonup }{J}d\tau$

Hence, the equation is proved.