A uniform line charge that has a linear charge density lambda = 2.7 nC/m is on t

Question

A uniform line charge that has a linear charge density lambda = 2.7 nC/m is on the x axis between x = 0 to x = 5.0 m. (a) What is its total charge? nC (b) Find the electric field on the x axis at x = 6 m. N/C (c) Find the electric field on the x axis at x = 12.0 m. N/C (d) Find the electric field on the x axis at x = 260 m. N/C (e) Estimate the electric field at x = 260 m, using the approximation that the charge is a point charge on the x axis at x = 2.5 m. N/C (f) Compare your result with the result calculated in part (d) by finding the ratio of the approximation to the exact result. To do this, you will need to assume that the values given in this problem statement are valid to more than two significant figures. (g) Is your approximate result greater or smaller than the exact result? greater smaller Explain your answer. (Do this on paper. Your instructor may ask you to turn in your work.)

Explanation / Answer

We can use the definition of to find the total charge of the line of charge and the expression for the electric field on the axis of a finite line of charge to evaluate Ex at the given locations along the x axis. In Part (d) we can apply Coulomb’s law for the electric field due to a point charge to approximate the electric field at x = 250 m

(a) Use the definition of linear charge density to express Q in terms of :

Q = L (2.7 nC/m)(5.0cm) =13.5 nC = [14 nC]

(a) Answer = 14 nC

Express the electric field on the axis of a finite line charge:

Ex(x0) = kQ / x0(x0 - L)

(b)Substitute numerical values and evaluate Ex at x = 6.0 m:

Ex (6.0m) = (8.988 * 10 ^9 N-m2/C2)(13.5nC) / (6.0 m)(6.0m - 5.0m) = 20N/C

(c) Substitute numerical values and evaluate Ex at x = 12.0 m

   Ex (6.0m) = (8.988 * 10 ^9 N-m2/C2)(13.5nC) / (12.0 m)(12.0m - 5.0m) = 1.44N/C

(d) Substitute numerical values and evaluate Ex at x = 260.0 m

  Ex (6.0m) = (8.988 * 10 ^9 N-m2/C2)(13.5nC) / (260.0 m)(260.0m - 5.0m) = 1.83mN/C

(e) Use Coulomb’s law for the electric field due to a point charge to obtain:

Ex = kQ / x^2

Substitute numerical values and evaluate Ex(260 m):

Ex(260m) = (8.988 * 10 ^9 N-m2/C2)(13.5nC) / ( 260 -2.5 )^2 = 1.82 mN/C

(f) This result is about 0.01% less than the exact value obtained in (d). This suggests that the line of charge is too long for its field at a distance of 260 m to be modeled exactly as that due to a point charge.

(g) The approximate result is smaller than the exact result.

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