Physics 230 Spring 2001
 
 

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 y4. And then E decreases like 1/y3 as we expected.