?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 5.80 nC. Calculate successive approximations for the electric potential at the origin by modeling the filament in the following ways.

A uniformly charged filament lies along the x axis between x = a = 1.00 m and x = a + 1 = 7.00 m as shown in the figure below. The total charge on the filament is 5.80 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 (b) as two 2.90 nC charged particles at x = 2.50 m and x = 5.50 m (c) as four 1.45 nC charged particles at x = 1.75 m, x = 3.25 m, x = 4.75 m, and x = 6.25 m (d) Explain how the results compare with the potential given by the exact expression () The exact result, represented as V,-Select the four-particle version. Modeling the line as a set of points Select

Explanation / Answer

a) potential due to a point charge, V = kq /r

V = (9 x 10^9 x 5.80 x 10^-9) / 4

V = 13.05 Volt

b)

V = V1 + V2

V = [ kq1/r1] + [kq2/r2]

V = [ ( 9 x 10^9x 2.90 x 10^-9 ) / (2.50) ] + [ ( 9 x 10^9x 2.90 x 10^-9 ) / (5.50) ]

V = 10.44 + 4.75 = 15.19 volt

c) V = [ ( 9 x 10^9x 1.45 x 10^-9 ) / (1.75) ] + [ ( 9 x 10^9x 1.45 x 10^-9 ) / (3.25 ]

+ [ ( 9 x 10^9x 1.45x 10^-9 ) / (4.75) ] + [ ( 9 x 10^9x 1.45 x 10^-9 ) / (6.25) ]

V = 16.31 volt

d) V = kq/l ln (l +a / a)

V = (9 x 10^9 x5.80 x 10^-9 / 6 ) ln ( 7/1)

V = 8.7 ln(7) = 16.93 volt

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