Question

An annular disc of inner radius R and Outer radius 2r has uniformly distributed charge Q electric field strength at point P is

Answer 1

Answer:

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.

Electric Field on the Axis of a Uniformly Charged Disk

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