Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First of all, we need to sketch the problem, where the dipole consists of two equla-unlike charges, as seen below.
From the geometry of the figure below, we can see that:
$\bullet$ $\theta_1=\theta_2=\theta$
$\bullet$ $r_1=r_2=\sqrt{x^2+(s/2)^2}$
$\bullet$ $\sin\theta=\dfrac{s/2}{r}=\dfrac{s}{2\sqrt{x^2+(s/2)^2}}$
$\bullet$ $\cos\theta=\dfrac{x}{r}=\dfrac{x}{\sqrt{x^2+(s/2)^2}}$
Thus, the electric field exerted by $q_1$ at a point at a distance of $x$, as seen below, is given by
$$E_1=\dfrac{kq_1}{r_1^2}\left(\cos\theta\;\hat i-\sin\theta\;\hat j\right)$$
Plug the known;
$$E_1=\dfrac{kq}{(x^2+(s/2)^2)}\left(\dfrac{x}{\sqrt{x^2+(s/2)^2}}\;\hat i-\dfrac{s}{2\sqrt{x^2+(s/2)^2}}\;\hat j\right)$$
$$E_1=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i-\dfrac{s}{2 }\;\hat j\right)\tag 1$$
The electric field by $q_2$ at the same point is given by
$$E_2=\dfrac{kq_2}{r_2^2}\left(-\cos\theta\;\hat i-\sin\theta\;\hat j\right)$$
Plug the known;
$$E_2=\dfrac{-kq}{(x^2+(s/2)^2)}\left(\dfrac{x}{\sqrt{x^2+(s/2)^2}}\;\hat i+\dfrac{s}{2\sqrt{x^2+(s/2)^2}}\;\hat j\right)$$
$$E_2=\dfrac{-kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i+\dfrac{s}{2 }\;\hat j\right)\tag 2$$
Thus, the net electric field is given by
$$E_{dipole}=E_1+E_2$$
Plug from (1) and (2),
$$E_{dipole}=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i-\dfrac{s}{2 }\;\hat j\right)-\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left(x\;\hat i+\dfrac{s}{2 }\;\hat j\right)$$
$$E_{dipole}=\dfrac{kq}{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}}\left[ x\;\hat i-\dfrac{s}{2 }\;\hat j -( x\;\hat i+\dfrac{s}{2 }\;\hat j)\right]$$
$$ E_{dipole}= \dfrac{ -kqs }{\left(x^2+\dfrac{s^2}{4}\right)^{3/2}} \;\hat j\tag 3$$
And when $x$ is much greater than the length of the dipole $s$, $(x\gt \gt s)$, then $\left(x^2+\dfrac{s^2}{4}\right)^{3/2}\approx x^3$,
Plug into (3),
$$ E_{dipole}= \dfrac{ -kqs }{x^3} \;\hat j $$
Note that:
- We can replace $r$ by $x$ for generalizing the term.
- Recall that $\vec p=qs$
- Recall that $k=1/4\pi \epsilon_0$
$$ \boxed{\vec E_{dipole}=- \dfrac{ \vec p }{4\pi \epsilon_0 r^3} } $$
Where the direction of the elipole electric field, in this case, is in the opposite direction of the electric dipole moment.
