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
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,
denominator is approximately equal to y4. And then E
decreases like 1/y3 as we expected.