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, ρ_b = −div($\vec P$). The second equation for the potential function can be written as the potential due to point dipoles $\vec P$dV_vol 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, V_vol, 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 V_vol 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.