In the figure a nonconducting rod of length L = 8.32 cm has charge -q = -4.35 fC

Question

In the figure a nonconducting rod of length L = 8.32 cm has charge -q = -4.35 fC uniformly distributed along its length, (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 14.8 cm from the rod? What is the electric field magnitude produced at distance a = 84 m by (d) the rod and (e) a particle of charge -q = -4.35 fC that replaces the rod?

Explanation / Answer

(a)

lamda = -q/L = -4.35 * 10^-15/0.0832 = -5.22 * 10^-14 C/m

(b)

Ex = - k q/ a ( L+a) = - 9* 10^9) ( 4.35 * 10^-15)/0.148 ( 0.0832+0.148) = -1.14 * 10^-3 N/C

magnitude = 1.14 * 10^-3 N/C

(c)

The negative sign in x E indicates that the field points in the –x direction, or 180
counterclockwise from the +x axis

(d)

If a is much larger than L, the quantity L + a in the denominator can be approximated
by a, and the expression for the electric field becomes

Ex = -kq/a^2

= - 9.0 * 10^9 ( 4.35 * 10^-15)/84^2

=-5.54 * 10^-9 N/C

magnitude 5.54 * 10^-9 N/C

(e)

E=5.54 * 10^-9 N/C

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