Undergrad The electric field due to a dipole

[wolly]

and

Can someone explain how they made these equations like this?
How did the radius become that equation?
What formulas from algebra did they applied?
I'm looking at these formulas and I don't understand how r=z+1/2*d

 

[Ibix]

The dipole charges are defined to be on the z axis at $z=\pm d/2$. If I point out that your textbook is not using full-on vector algebra, so is working in 1d, can you see how to get from $r^2$ to $(z\pm d/2)^2$?

 

[vanhees71]

From this little example, I can only come to the conclusion to recommend to use another book. SCNR.

The idea of the dipole is to have a charge $Q$ at $(0,0,d/2)=d/2 \vec{e}_3$ and $-Q$ at $(0,0,-d/2)=-d/2 \vec{e}_3$. The field is given by the superposition of the field of these point charges (Coulomb fields), i.e.,
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0} \left [\frac{\vec{r}-d/2 \vec{e}_3}{|(\vec{r}-d/2 \vec{e}_3|^3}-\frac{\vec{r}+d/2 \vec{e}_3}{|\vec{r}+d/2 \vec{e}_3|^3} \right].$$
The next step usually is to make $d \rightarrow 0$ while keeping $Q d=\text{const}$. This leads to the dipole approximation of the field for this charge configuration. This is a somewhat cumbersome calculation, however.

It's easier to use the electrostatic potential:
$$\Phi(\vec{r},d)=\frac{Q}{4 \pi \epsilon_0} \left [\frac{1}{|\vec{r}-d/2 \vec{e}_3|} - \frac{1}{|\vec{r}+d/2 \vec{e}_3|} \right].$$
Now $|\vec{r} \pm d/2 \vec{e}_3|=\sqrt{x_1^2 +x_2^2 + (x_3\pm d/2)}^2$.
The Taylor expansion of $\Phi$ wrt. $d$ is
$$\Phi(\vec{r},d)=\Phi(\vec{r},0) + d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2) = d \partial_d \Phi(\vec{r},0)+\mathcal{O}(d^2).$$
This gives
$$\Phi(\vec{r},d)=\frac{Q d x_3}{4 \pi \epsilon_0 r^3} + \mathcal{O}(Q d^2).$$
Defining the dipole moment
$$\vec{p}=Q d \vec{e}_3$$
Making now $d \rightarrow 0$ but keeping $Q d=\text{const}$ you get the dipole potential,
$$\Phi(\vec{r})=\frac{\vec{p} \cdot \vec{r}}{4 \pi \epsilon r^3}.$$
The field is
$$\vec{E}=-\vec{\nabla} \Phi=-\frac{\vec{p}}{4 \pi \epsilon_0 r^3} + \frac{3 (\vec{p} \cdot \vec{r}) \vec{r}}{4 \pi \epsilon_0 r^5} = \frac{1}{4 \pi \epsilon_0 r^5} [3 (\vec{r} \cdot \vec{p}) \vec{r}-r^2 \vec{p}].$$