What is the magnitude of the electric field at the center of a ring of charge of radius a? Assume there is a charge Q uniformly distributed over the ring.
The field from one side of the ring cancels the field from the other, so the net field at the center is zero.
What is the magnitude of the electric field a distance x from the center of the ring, along the axis of the ring?
The field from one part of the ring is partly cancelled by the field from the opposite side of the ring. The field is nonzero but has a magnitude less than kQ/r^{2}. The field can be determined exactly by integrating.
The strategy is to split the ring up into tiny pieces and give each piece a tiny charge dQ. Each of these charges sets up a tiny field dE at our point, where:
dE  = 

Add these fields as vectors to get the net field:
E = Σ dE
When we integrate to find the net field all the components along the axis add and all the components perpendicular to the axis cancel. We put a factor of:
cos(θ) = x/r
in the integral so we're only including the components along the axis.
E = E_{x}  =  ∫ 

=  ∫ 

For every piece of the ring r^{2} = x^{2} + a^{2}
E = E_{x}  = 

ò  dQ  = 

One way to check your answer is to plug in the extremes. When x is zero, which means the point is at the center of the ring, the equation gives E = 0, which is good. When x is much larger than a the expression reduces to the point charge equation. This also makes sense, seeing as when you're very far away from the ring it looks like a point charge.