I am a bit lost how one has deduced the formula for electric field with electric dipole because of some inconsistency between different sources. The Wikipedia article contains a delta function in the formula, but this paper gives only the formula without the delta function,
$$\bar{E} = -\nabla \phi = \frac{-p}{4\pi \epsilon_{0} r^{3}} ( \hat{e}_{z}-3\hat{u}_{r}(\hat{u}_{r} \cdot \hat{u}_{z}) ) $$
Is it intentionally left out?
My notes mention that the dipole is with a circular thing $P=Q d$ where $d$ is distance and $Q$ is charge.
Answer
I'll give you the derivation from my book which includes a nice way to see how the delta functions arise: ..............................................................................................................................................................
We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and $\vec{B}$ of a vector point dipole and an axial point dipole in the same way as we derive these in the case of a the point monopole. We start with a static point charge $\delta(\vec{r})$ and derive the dipole charge/current densities with the help of differential operators.
We apply the same differential operators for $\vec{A}$, $\vec{E}$ and $\vec{B}$ to obtain the dipole fields from the monopole fields. First we recall the fields of the monopole. The point charge obtains (over time) a potential field given by.
$ \mbox{field}\Big\{\,\delta(\vec{r})\,\Big\} ~~=~~ \frac{1}{4\pi r} $
The reversed operator which derives the source from the field is just the Laplacian operator. $ -\nabla^2\Big\{\,\frac{1}{4\pi r}\,\Big\} ~~=~~ \delta(\vec{r}) $
Integrating the delta function over space shows equal contributions from the three spatial components.
$ \int \delta(\vec{r})~d\vec{r} ~=~ -\int\left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right] \frac{d\vec{r}}{4\pi r} ~=~ \frac13+\frac13+\frac13 ~=~ 1 $
We can define vector dipole and axial dipole sources by using differential operators on the monopole $\delta(\vec{r})$ and derive their potential and electromagnetic fields.
The vector dipole is obtained by differentiating the monopole along the direction that we want the dipole to have. The result is a combination of a positive and a negative delta function. The axial dipole is defined by the curl of the monopole so that it gives a circular point current in the direction of the dipole.
charge/current densities, potentials and fields of dipoles:
$ \begin{array}{|lcll|} \hline &&& \\ j^o &=& ~~~\mbox{div}\,\left(~\vec{\mu} \,\delta(\vec{r})~\right) & \mbox{vector dipole charge density} \\ \vec{j} &=& ~~~\mbox{curl}\left(~\vec{\mu}\, \delta(\vec{r})~\right) & \mbox{axial dipole current density} \\ &&& \\ A^o &=& ~~~\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{vector dipole electric potential} \\ \vec{A} &=& ~~~\mbox{curl}\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right) & \mbox{axial dipole magnetic potential} \\ &&& \\ \mathsf{E}&=& -~\mbox{grad}\left(\mbox{div}\,\left(~\vec{\mu} \,\frac{1}{4\pi r}\,\right)\right)& \mbox{vector dipole electric field} \\ \mathsf{B}&=& +~\,\mbox{curl}\left(\mbox{curl} \left(~\vec{\mu}\,\frac{1}{4\pi r}\,\right)\right)& \mbox{axial dipole magnetic field} \\ &&& \\ \hline \end{array} $
The potential fields are obtained by applying the same differential operators on the field $1/4\pi r$ of the monopole rather than on the charge distribution. For a dipole moment in the $z$-direction we obtain in SI-units.
Potential fields of the electric and magnetic dipoles:
$\Phi ~=~ \frac{z}{4\pi\epsilon_o r^3}\,\mu~~~~~~~~ \vec{A} ~=~ \Big( -y ~,~ x~ ~,~ 0 \Big)~ \frac{\mu_o}{4\pi r^3}\,\mu $
The expressions for the electromagnetic fields do implicitly contain delta functions at the center with the right magnitude. These delta functions are easily lost if the derivation isn't careful enough. The E and B fields are related to each other by the standard vector identity.
$\mbox{curl}(\mbox{curl}\vec{X}) ~=~ \mbox{grad}(\mbox{div}\vec{X})-\nabla^2\vec{X} $
We have seen that the last term (the Laplacian) yields $\delta(\vec{r})$. The two therefor differ only at the center by a delta function in the $\vec{\mu}$ direction. We can see this explicitly if we align the dipoles with the z-axis, so that $\vec{\mu}$ has only a z-component, and write out the fields.
The only difference is in the z-components. The total difference between the two is the Laplacian and thus $\delta(\vec{r})$. The vector dipole gets -1/3 while the axial dipole gets +2/3.
$\begin{aligned} \mathsf{E} &= &\frac{1}{\epsilon_o}\bigg[~ \mathbf{\hat{x}}\, \frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\, \frac{\partial}{\partial y}\frac{\partial}{\partial z} ~+~~~ \mathbf{\hat{z}}\, \frac{\partial^2}{\partial z^2} ~~~~~~~~~~~~~ &\bigg] ~ \frac{\mu}{4\pi r}\\ \\ \mathsf{B} &= &\mu_o\bigg[~ \mathbf{\hat{x}}\,\frac{\partial}{\partial x}\frac{\partial}{\partial z} ~+~ \mathbf{\hat{y}}\,\frac{\partial}{\partial y}\frac{\partial}{\partial z} ~-~ \mathbf{\hat{z}}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)~ &\bigg] ~ \frac{\mu}{4\pi r}\\ \end{aligned} $
Which we can see from the third equation with the three 1/3 parts. In the general case, with arbitrary dipole direction $\vec{\mu}$ we get for the electromagnetic fields in vector form.
Electromagnetic dipole fields:
$ \begin{aligned} \mathsf{E} &= &\frac{1}{\epsilon_o}\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~-~\frac13\vec{\mu}\,\delta(\vec{r})~\right) \\ \\ \mathsf{B} &= &\mu_o\left(~\frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} ~+~\frac23\vec{\mu}\,\delta(\vec{r})~\right) \end{aligned} $
Alternatively we can look at the fields explicitly expressed in the individual $x$, $y$ and $z$ components which gives. (with the dipole moment in the $z$-direction)
$\begin{aligned} \mathsf{E} &= &\frac{\mu}{4\pi\epsilon_o}\bigg[~ \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~ \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~ \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} - \frac{4\pi}{3}\delta(r)\right)~ &\bigg] \\ \\ \mathsf{B} &= &\frac{\mu_o\mu}{~4\pi~}\bigg[~ \mathbf{\hat{x}}\, \frac{3xz}{r^5} ~+~ \mathbf{\hat{y}}\, \frac{3yz}{r^5} ~+~~~ \mathbf{\hat{z}}\, \left(\frac{3zz}{r^5}-\frac{1}{r^3} + \frac{8\pi}{3}\delta(r)\right)~ &\bigg] \\ \end{aligned} $
In general the fields decrease with the third order of $r$ compared to $1/r^2$ for the charge which makes it possible not many effects at larger scale. At the other hand, when we go to smaller scales, the magnetic dipole fields become more powerful relative to the charge.
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