V. A charge of -6.50 nonconducting disk of radius 1.25 cm. A.Find the magnitude

Question

V. A charge of -6.50 nonconducting disk of radius 1.25 cm. A.Find the magnitude and direction of the electric field this disk produces at a point P on the axis of the disk a distance of 2.00 cm from its center nC is spread uniformly over the surface of one face of a C. Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude and direction of the electric field at point P. D. If the charge is all brought to the center of the disk, find the magnitude and direction of the electric field at point P. E. why is the field in part A stronger than the field in part B? Why is the field in part C the strongest of the three fields?

Explanation / Answer

(A) field due to charged disc,

E = (2 pi sigma k )[1 - x/sqrt(x^2 + R^2)]

{ 2 pi sigma k = 2 pi (Q / pi r^2) k = 2 k Q / R^2}

E = (2 x 6.50 x 10^-9 x 9 x 10^9/0.0125^2)(1 - (2/sqrt(2^2 + 1.25^2)))

E = 1.14 x 10^5 N/C

direction-> toward the center of the disc

(c) E = k Q x / (x^2 + R^2)^(3/2)

E = (9 x 10^9)(6.50 x 10^-9)(0.02)/(0.02 + 0.0125)^(3/2)

E = 8.92 x 10^4 N/C

(D) E = k Q / R^2 = 3.74 x 10^5 N/C

(E) in part(C), all the charges are at closest distance possible so strongest field strength.

in B , charges are at furthest distance possible.

hence weakest field strength.

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