In the figure a nonconducting rod of length L = 8.34 cm has charge - q = -4.31 f

Question

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.31 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 = 14.7 cm from the rod? What is the electric field magnitude produced at distance a = 74 m by (d) the rod and (e) a particle of charge -q = -4.31 fC that replaces the rod?

Explanation / Answer

charge density = Q/L = 4.31*10^-15/(8.34*10^-2) = 5.17*10^-14 C/m

(b)

let p be the origin

consider a small charge dq at x from p

dq = (Q/L)*dx

dE = k*dq/(x)^2

dE = (k*Q/L)*(dx/x^2)

E = integration dE from x = a to x = L+a

E = (k*Q/L)*(-1/x) from x = a to x = L+a

E = (kQ/L)*( 1/a - 1/(L+a) )

for a = 14.7 = 0.14 m

E = (9*10^9*4.31*10^-15/0.0834)*(1/0.147 - 1/(0.0834+0.147))

E = 0.00114 N/C

-----------------

direction towards ledf +180 degrees

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part(d)

for a = 74 m

E = (9*10^9*4.31*10^-15/0.0834)*(1/74 - 1/(0.0834+74))

E = 7.08*10^-9 N/C

(e)

E = k*q/a^2 = 9*10^9*4.31*10^-15/74^2

E = 7.08*10^-9 N/C

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