In the figure a nonconducting rod of length L = 8.37 cm has charge -q = -4.36 fC

Question

In the figure a nonconducting rod of length L = 8.37 cm has charge -q = -4.36 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 = 13.6 cm from the rod? What is the electric field magnitude produced at distance a = 67 m by (d) the rod and (e) a particle of charge -q = -4.36 fC that replaces the rod? Number Number Number Number Number Units Units Units Units Units

Explanation / Answer

(a)

linear charge density = q/L = (4.36*10^-15)/0.0837 = 5.2*10^-14 C/m

(b)

consider a small length dx at distance x from P

small chagrge dq = (q/L)*dx

small electric field due to dx , dE = -k*dq/x^2

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

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

total electric field E = integral dE from x = a to x = a + L

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

E = -(k*q/L)*( -1/(L+a) + 1/a )

(b)

magnitude E = (k*q/L)*( 1/a -1/(L+a) )

E = (9*10^9*4.36*10^-15/0.0837)*(1/0.136 - 1/(0.0837+0.136))

E = 0.00131 N/C

(c)

directon = -x axis

(d)

E = (9*10^9*4.36*10^-15/0.0837)*(1/67 - 1/(0.0837+67))

E = 8.73*10^-9 N/C

(e)

E = k*q/a^2 = 9*10^9*4.36*10^-15/67^2 = 8.73*10^-9 N/C

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