Find the Expression of an Electric Field Produced by a Dipole

[ladyrx2020]

Homework Statement

An electric dipole is located along the y-axis as shown in (picture attached). The magnitude of its electric dipole moment is defined as $p=2aq$.

(a) At a point P, which is far from the dipole $(r>>a)$, show that the electric potential is
$V=\frac{k_{e}pcosθ}{r^{2}}$

(b) Find an expression of the electric field produced by the dipole in terms of the vectors $\hat{r}$ and $\vec{p}$ (the dipole moment).

Homework Equations

$p=2aq$ (Magnitude of the Electric Dipole Moment)
$\vec{E}=k_{e}\frac{q}{r^{2}}\hat{r}$ (Electric Field at P created by a q charge)

The Attempt at a Solution

Part (a):
$V=\frac{k_{e}q}{r_{1}}-\frac{k_{e}q}{r_{2}}=\frac{k_{e}q}{r_{1}r_{2}}(r_{2}-r_{1})$
Since $r>>a$ and $r_{2}-r_{1}≈2acosθ$, then $v≈\frac{k_{e}q}{r_{1}r_{2}}2acosθ≈\frac{k_{e}pcosθ}{r^{2}}$

Part (b):
I need help with this part.

So far, I got that both +q and -q is $E=k_{e}\frac{|q|}{r^{2}}$
Not sure how to complete the rest since point P does not lie on the x-axis.

NOTE:
An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance of 2a. The electric dipole moment $\vec{p}$ is directed from -q toward +q.

 

[BruceW]

in part a) what happened to r1 in the denominator? p.s. I think you have written down the electric potential V incorrectly, it should not be:
$$V=\frac{k_{e}pcosθ}{r_{2}}$$
And for part b), you have written down the correct equation for the magnitude of the electric field due to a point particle, but what you need is the direction as well!

 

[vanhees71]

The idea is to perform a Taylor expansion.

The full expression for the potential reads
$$V(\vec{r})=q k_e \left (\frac{1}{\sqrt{(\vec{r}-a \vec{e}_y)^2}}-\frac{1}{\sqrt{(\vec{r}+a \vec{e}_y)^2}} \right ).$$
The point is that $a/|\vec{r}|\ ll 1$ if you look at the field from a distance far from the charges. Then in leading non-vanishing order in the expansion in powers of $a/|\vec{r}|$ you find
$$V(\vec{r})=q k_e \left [-2a \vec{e}_y \cdot \vec{\nabla} \frac{1}{r} \right ]=k_e \frac{\vec{p} \cdot \vec{r}}{r^3}.$$

Note that the expression given in your posting cannot be correct for dimensional reasons!

You find the electric field by taking the gradient of the potential
$$\vec{E}=-\vec{\nabla} V(\vec{r}).$$