In the figure below, a nonconducting rod of length L = 8.50 cm has charge -q = -

Question

In the figure below, a nonconducting rod of length L = 8.50 cm has charge -q = -4.24 fC uniformly distributed along its length.

http://www.webassign.net/hrw/hrw7_22-43.gif


(a) What is the linear charge density of the rod?
C/m

(b) What is the magnitude of the electric field at point P, a distance a = 12.0 cm from the end of the rod?
N/C

(c) What is its direction?
° (counterclockwise from the positive x axis)

(d) What is the electric field magnitude produced at distance a = 50 m by the rod?
N/C

(e) Repeat part (d) for a particle of charge -q = -4.24 fC that replaces the rod.
N/C

Explanation / Answer

a) lampa = q / L q = -4.24 fC = -4.24*10^-15 C L = 8.50 cm => 0.085 m lampa = -4.988 * 10 ^ -14 C/m b) lampa * K [ 1/ (0.205 -0.085) - (1/0.205) ] = 4 . 489 * 10 ^ -4 (8.333 - 4.878) = 0.00155 N / C {POSitive ANS} c) theta = 180 ° d) E(x) = - K*q / a^2 a = 50 m so E(x) = 1.526 * 18 ^ -8 N/C e) will be the same as d E(x) = 1.526 * 18 ^ -8 N/C

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