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 5.00R 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/5.00 (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 at the axial point for a charged disc is p/2e * ( 1 - x/(x^2+R^2)^1/2 )

Where p= surface charge density

e= 8.85*10^-12

x= axial distance and R= radius

So, for the first case, the electric field at the point P is p/2e * ( 1 - 5R/(25R^2+R^2)^1/2 ) = 0.0194*p/2e

And in the second case, the ring is equivalent to the disc of radius 0.2R removed from the previous disc. So the electric field at point P is 0.0194p/2e - [ p/2e * ( 1 - 5R/(25R^2+0.04R^2) ] = 0.0186p/2e

So, 0.0194p/2e -0.0186p/2e / 0.0194p/2e = 4/97 . So, 4/97 of the electric field magnitude is decreased at point P.

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