Find the electric field at a point on the *y*-axis with *y >
d*/2 due to two point charges ± *q*
on the *y*-axis at *y = - d*/2 (the
positive charge) and *y = + d*/2 (the negative
charge).

Here's the diagram:

The two electric field vectors
lie along the *y*-axis. The field due to the negative charge has the
larger magnitude because point P is closer to the negative charge than
to the positive charge. Thus the sum of the two vectors is in the minus-*y*
direction. Its magnitude is the difference of the two magnitudes.

The magnitudes are given by Coulomb's law. The distance from P to the
charge -q is *y-d*/2, so

while for the positive charge we have

so the total field is:

So when *y* >> *d,
*the
denominator is approximately equal to *y*^{4}. And then E
decreases like 1/*y*^{3} as we expected.