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 2.40Rfrom 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/2.40 (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

a)Electric field due to disk=2pi*k*sigma*(1-x/sqrt(x^2+R^2))

Field due to ring=field due to outer disc-field due to inner disc=2pi*k*sigma*(1-x/sqrt(x^2+R^2))-2pi*k*sigma*(1-x/sqrt(x^2+(R/2.4)^2))

Fraction loss of field=2pi*k*sigma*(1-x/sqrt(x^2+R^2))-(2pi*k*sigma*(1-x/sqrt(x^2+R^2))-2pi*k*sigma*(1-x/sqrt(x^2+(R/2.4)^2)))/(2pi*k*sigma*(1-x/sqrt(x^2+R^2)))

=2pi*k*sigma*(1-x/sqrt(x^2+(R/2.4)^2))/(2pi*k*sigma*(1-x/sqrt(x^2+R^2))=(1-x/sqrt(x^2+(R/2.4)^2))/(1-x/sqrt(x^2+R^2))

Here x=2.4R

Fraction loss=(1-2.4R/sqrt((2.4R)^2+(R/2.4)^2))/(1-2.4R/sqrt((2.4R)^2+R^2))=0.1916

The loss is only 19.16%

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