Question
Report part(a), only this time make the right hand charge -q instead of +q.
Answer
As we can easily conclude from the diagram, vertical components here cancel each other, only surviving components are horizontals. Consider this two dimensional problem is confined in the z-x plane.
So the horizontal resultant force field along x-axis is,
$E_x = 2q sin\theta/4\pi\epsilon_0 r^2$
From the diagram,
$sin\theta = d/2r $
and $r = \sqrt{z^2 + (d/2)^2}$
Therefore net force field is as follows,
$E = qd\hat{x}/4\pi\epsilon_0 (z^2 + (d/2)^2)^{3/2}$
For z>> d the field goes like,
$E = qd\hat{x}/4\pi\epsilon_0 z^3$
This is as we know is the field of a dipole.
Report part(a), only this time make the right hand charge -q instead of +q.
Answer
So the horizontal resultant force field along x-axis is,
$E_x = 2q sin\theta/4\pi\epsilon_0 r^2$
From the diagram,
$sin\theta = d/2r $
and $r = \sqrt{z^2 + (d/2)^2}$
Therefore net force field is as follows,
$E = qd\hat{x}/4\pi\epsilon_0 (z^2 + (d/2)^2)^{3/2}$
For z>> d the field goes like,
$E = qd\hat{x}/4\pi\epsilon_0 z^3$
This is as we know is the field of a dipole.