EM1_L3

We know that the electric field at $\mathbf{r}$ is given by

$$\begin{array}{r}\mathbf{E}(\mathbf{r})=k\int \rho ({\mathbf{r}}^{\prime })\frac{\mathbf{r}-{\mathbf{r}}^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}{d}^3{\mathbf{r}}^{\prime }\end{array}$$

The $p$th component of the electric field is given by

$$\begin{array}{r}{E}_p(\mathbf{r})=k\int \rho ({\mathbf{r}}^{\prime })\frac{r_p-r_p^{\prime }}{|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3}{d}^3{\mathbf{r}}^{\prime }\end{array}$$

Note that

$$\begin{array}{rlr}\frac{\partial }{\partial x_p^{\prime }}\left[\frac{1}{|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^n}\right]& =\frac{\partial }{\partial x_p^{\prime }}\left[\frac{1}{|(x_j-x_j^{\prime }{)}^2{|}^{n/2}}\right]& \\ & =-\frac{n}{2}\frac{\left(\frac{\partial }{\partial x_p^{\prime }}((x_j-x_j^{\prime }{)}^2)\right)}{|(x_j-x_j^{\prime }{)}^2{|}^{n/2+1}}& \\ & =\frac{n(x_p-x_p^{\prime })}{|\mathbf{x}-{\mathbf{x}}^{\prime }{|}^{n+2}}& (1)\end{array}$$

We get a nice substitution for $n=1$:

$$\begin{array}{rl}{E}_p(\mathbf{r})& =k\int \rho ({\mathbf{r}}^{\prime })\frac{\partial }{\partial r_p^{\prime }}\left[\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\right]d^3{\mathbf{r}}^{\prime }\end{array}$$

Observe from $(1)$ that

$$\frac{\partial }{\partial x_p^{\prime }}\left[\frac{1}{|\mathbf{x}-{\mathbf{x}}^{\prime }|}\right]=-\frac{\partial }{\partial x_p}\left[\frac{1}{|\mathbf{x}-{\mathbf{x}}^{\prime }|}\right]$$

So,

$$\begin{array}{rl}{E}_p(\mathbf{r})& =-k\frac{\partial }{\partial r_p}\int \rho ({\mathbf{r}}^{\prime })\left[\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\right]d^3{\mathbf{r}}^{\prime }\end{array}$$

Define

$$\varphi (\mathbf{r})\equiv k\int \rho ({\mathbf{r}}^{\prime })\left[\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\right]d^3{\mathbf{r}}^{\prime }$$

Then,

$$E_p(\mathbf{r})=-\frac{\partial \varphi }{\partial r_p}$$ $$\mathbf{E}(\mathbf{r})=-\nabla \varphi$$

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Example: point charge
$\rho ({\mathbf{r}}^{\prime })=Q\delta ({\mathbf{r}}^{\prime }-\mathbf{a})$

$$\varphi ({\mathbf{r}}^{\prime })=\frac{kQ}{|{\mathbf{r}}^{\prime }-\mathbf{a}|}+c$$

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Example: dipole
$\rho ({\mathbf{r}}^{\prime })=-\mathbf{p}\cdot \nabla {\delta }^3({\mathbf{r}}^{\prime }-\mathbf{a})$

$$\begin{array}{rl}\varphi (\mathbf{r})& =k\int \frac{-\mathbf{p}\cdot \nabla {\delta }^3({\mathbf{r}}^{\prime }-\mathbf{a})}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}{d}^3{\mathbf{r}}^{\prime }\\ & =-k{p}_i\int \left(\frac{\partial }{\partial r_i^{\prime }}{\delta }^3({\mathbf{r}}^{\prime }-\mathbf{a})\right)\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}{d}^3{\mathbf{r}}^{\prime }\\ & =k{p}_i\int \left[\frac{\partial }{\partial r_i^{\prime }}\frac{1}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}\right]{\delta }^3({\mathbf{r}}^{\prime }-\mathbf{a})d^3{\mathbf{r}}^{\prime }\end{array}$$