A uniformly charged filament lies along the x axis between x = a = 1.00 m and x

Question

A uniformly charged filament lies along the x axis between x = a = 1.00 m and x = a + ? = 7.00 m as shown in the figure below. The total charge on the filament is 4.50 nC. Calculate successive approximations for the electric potential at the origin by modeling the filament in the following ways.

(a) as a single charged particle at x = 4.00 m
________V

(b) as two 2.25 nC charged particles at x = 2.50 m and x = 5.50 m
________V

(c) as four 1.13 nC charged particles at x = 1.75 m, x = 3.25 m, x = 4.75 m, and x = 6.25 m
________V

(d) Explain how the results compare with the potential given by the exact expression

The exact result, represented as ________V,  ---Select--- is approximated to within 5% by or differs by between 5 and 10% from differs by greater than 10% from the four-particle version. Modeling the line as a set of points  ---Select--- is a close approximation or is not a good approximation.

Explanation / Answer

a) V = kq/r = 9 x 109 x 4.5 x 10-9/4 = 10.125 V

b) V = 9 x 109 x 2.25 x 10-9 x [(1/2.5) + (1/5.5)] = 11.782 V

c) V = 9 x 109 x 1.13 x 10-9 x [(1/1.75) + (1/3.25) + (1/4.75) + (1/6.25)] = 12.709 V

d) Vexact = kQ/L x ln[(L+a)/a] = (9 x 109 x 4.5 x 10-9/6) x ln7 = 13.135 V

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