*
[Note from ghw: This is a local copy of a portion of Stephen Kevan's lecture on Electric Fields and Charge Distribution of April 8, 1996.]*

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.

*
[Note from ghw: This is a local copy of a portion of Stephen Kevan's lecture on Electric Fields and Charge Distribution of April 8, 1996.]*

### Using the above result, we can easily derive the electric field on
the axis of a uniformly charged disk, simply by invoking superposition
and summing up contributions of a continuous distribution of rings, as
shown in the following figure from Tipler:

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.

## Electric Field near an Infinite Plane of Uniform Charge Density

### A much more important limit of the above result is actually for x much
less than R. In this case, it is as though the disk were of infinite extent,
so the result corresponds simply to the electric field near an infinite
sheet of charge. If we let R go to infinity (or at least to become very
large compared to x) we get the very simple result that

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