We determine the field at point P on the axis of the ring. It should be apparent from symmetry that the field is along the axis. The field dE due to a charge element dq is shown, and the total field is just the superposition of all such fields due to all charge elements around the ring. The perpendicular fields sum to zero, while the differential x-component of the field is
We now integrate, noting that r and x are constant for all points on the ring:
This gives the predicted result. Note that for x much larger than a (the radius of the ring), this reduces to a simple Coulomb field. This must happen since the ring looks like a point as we go far away from it. Also, as was the case for the gravitational field, this field has extrema at x = +/-a.
Such a surface charge density is conventionally given the symbol sigma. For a disk, we have the relationship
where Q is the total charge and R is the radius of the disk. A ring of thickness da centered on the disk as shown has differential area given by
and thus a charge given by
The field produced by this ring of charge is along the x-axis and is given by the previous result:
The total field is given by simply integrating over a from 0 to R
The integral is actually 'perfect' and is given by
After substituting the limits, we get the final result:
Very far from the disk, we need to use a series approximation with x much larger than R. The algebra is in Tipler, but rest assured that we simply recover the simple Coulomb law result.
This is a remarkable and useful result. For an infinite plane of charge, the field does not depend on x - we have a uniform field. If we can just figure out how to get a uniform plane of charge, then we can make an electron gun work. Recall that for an infinite line charge, the field decays as 1/r, while for a point charge it decays as 1/r^2. Anyone see a trend?
Now recall that field lines are directed away from positive charges and toward negative charges. A little thought will convince you that, if the charge density is positive (negative), the field must point away from (toward) the plane of charge on both sides! Thus, we have that
and there is a discontinuity in the electric field by 4*pi*k*sigma as we go through the plane.
Physics 253 Lecture Notes