Suppose you design an apparatus in which a uniformly charged disk of radius R is

Question

Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 3.90R from the disk (see Figure (a)). Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/3.90 (see Figure (b)). Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what part will you decrease the electric field magnitude at P?

Explanation / Answer

Electric Field on the Axis of a Ring of Charge [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. Electric Field on the Axis of a Uniformly Charged Disk [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:

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