Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges, q, a distance d apart (as shown in the figure). Check that your result is consistent with what you'd expect when z>>d.
Answer
As it is given in the problem, q charges are of same polarity. Naturally the directions of the force fields are as shown in figure, along AE and BD. It is clear from the figure that the horizontal components (hc) of the force fields cancel each other out, only surviving components are verticals. They are directed along the same direction. So the net force field is the simple algebraic sum of the two vertical components.
Therefore the resultant force field is as follows,
Therefore the resultant force field is as follows,
$E_z = {2q cos \theta}/{4\pi \epsilon_0 r^2}$
The factor 2 comes from the fact that, vertical forces are also equal in magnitude.
Here,
Here,
$r = \sqrt {z^2+(d/2)^2}$
and
$ cos\theta = sin (90-\theta) = z/r $
As we can see in the given figure.
So,
When z>>d, field point is so far away that from it the source charges almost look like a point charge of magnitude 2q.
Now the net field is,
So,
$E= 2qz\hat{z}/4\pi\epsilon_0 (z^2+(d/2)^2)^{3/2} $
When z>>d, field point is so far away that from it the source charges almost look like a point charge of magnitude 2q.
Now the net field is,
$E= 2qz\hat{z}/4\pi\epsilon_0 z^3 $
$E= 2q\hat{z}/4\pi\epsilon_0 z^2 $
We just set,
$ d \xrightarrow{} 0 $