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# 2.8: Appendix 2A

## 2.8 Appendix 2A.

It was pointed out in sections 2.2.1 and 2.2.2 above that the potential function, V($$\vec R$$) generated by a distribution of electric dipoles, P($$\vec r$$), can be calculated in two ways:

$\text { (1) } \quad \mathrm{V}(\overrightarrow{\mathrm{R}})=\frac{1}{4 \pi \epsilon_{0}} \int \int \int_{S_{\text {pace }} d V_{\text {vol }}} \frac{(-\operatorname{div}(\overrightarrow{\mathrm{P}}))}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}. \label{2.36}$

This equation for the potential function is calculated from the distribution of bound charges, ρb = −div($$\vec P$$). The second equation for the potential function can be written as the potential due to point dipoles $$\vec P$$dVvol summed over the entire distribution of dipoles:

$\text { (2) } \quad \mathrm{V}(\overrightarrow{\mathrm{R}})=\frac{1}{4 \pi \epsilon_{0}} \int \int \int_{S_{p a c e}} d V_{\text {vol }} \frac{\overrightarrow{\mathrm{P}} \cdot(\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|^{3}}. \label{2.37}$

These two formulae, Equations (\ref{2.36} and \ref{2.37}), give the same potential function apart from a possible constant that has no effect on the resulting electric field. This statement can be proved by applying Gauss’ Theorem to the function

$\operatorname{div}\left(\frac{\overrightarrow{\mathrm{P}}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right)=\operatorname{div}\left(\frac{\overrightarrow{\mathrm{P}}}{\sqrt{[X-x]^{2}+[Y-y]^{2}+[Z-z]^{2}}}\right). \nonumber$

The divergence is calculated with respect to the co-ordinates of the source point, (x,y,z):

$\operatorname{div}\left(\frac{\overrightarrow{\mathrm{P}}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right)=\frac{\partial}{\partial x}\left(\frac{\mathrm{P}_{x}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right)+\frac{\partial}{\partial y}\left(\frac{\mathrm{P}_{y}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right)+\frac{\partial}{\partial z}\left(\frac{\mathrm{P}_{z}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right).\nonumber$

By direct differentiation one can readily show that

$\operatorname{div}\left(\frac{\overrightarrow{\mathrm{P}}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}\right)=\frac{\operatorname{div}(\overrightarrow{\mathrm{P}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}+\frac{\overrightarrow{\mathrm{P}} \cdot(\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|^{3}}.\nonumber$

Remember that the differentiations are with respect to the co-ordinates of $$\vec r$$, (x,y,z), and not with respect to the observer co-ordinates $$\vec R$$, (X,Y,Z). Integrate the above equation over a volume, Vvol, bounded by a surface S and apply Gauss’ Theorem, section 1.3.3, to the term on the left. The result is

$\int \int_{S} \frac{d S(\overrightarrow{\mathrm{P}} \cdot \hat{\mathbf{n}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}=\int \int \int_{V_{\text {rot }}} \frac{d V_{\text {vol }} d i v(\overrightarrow{\mathrm{P}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}+\int \int \int_{V_{\text {rot }}} \frac{d V_{\text {vol }} \overrightarrow{\mathrm{P}} \cdot(\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|^{3}}.\nonumber$

Now let the volume Vvol become very large so that the surface S recedes to infinity. If the polarization distribution is limited to a finite region of space, as we shall assume, the surface integral must vanish because the polarization density on the surface, S, is zero. We are left with the identity

$-\int \int \int_{V_{\text {vol }}} \frac{d V_{\text {vol }} d i v(\overrightarrow{\mathrm{P}})}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|}=\int \int \int_{V_{\text {red }}} \frac{d V_{\text {vol }}}{|\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}|^{3}} \cdot(\overrightarrow{\mathrm{R}}-\overrightarrow{\mathrm{r}}) \label{2.38}$

Upon multiplying both sides of Equation (\ref{2.38}) by $$1 /\left(4 \pi \epsilon_{0}\right)$$ one obtains the integral of Equation (\ref{2.36}) on the left and the integral of Equation (\ref{2.37}) on the right. It follows that the same value for the potential will be obtained, aside from a possible unimportant constant, whether one uses the formulation based upon the potential for a point charge or the formulation based upon the potential function for a point dipole.