A charged line like that shown in the Fig.21.25 in the textbook extends from y =

Question

A charged line like that shown in the Fig.21.25 in the textbook extends from y = 2.50 cm to y = -2.50 cm. The total charge distributed uniformly along the line is -7.0 N/C.
a. Find the magnitude of the electric field on the x-axis at x = 10.0cm.
b.Find the direction of the electric field on the x-axis at x = 10.0 cm.
c. Is the magnitude of the electric field you calculated in part A larger or smaller than the electric field 10.0 from a point charge that has the same total charge as this finite line of charge?
d. At what distance does the result for the finite line of charge differ by 1.0% from that for the point charge?

Explanation / Answer

let,

y=2.5 cm

charge q=-7nC

x=10cm

a)

the eletric field due to finite charge distibution along the finite length is,

E=k*q/(x*sqrt(x^2+y^2))

E=(9*10^9)*(7*10^-9)/((10*10^-2)*sqrt((10*10^-2)^2+(2.5*10^-2)^2))

E=6111.9 N/C

b)

here, direction of electric field is toward the -x axis( or towards the finite wire)

C)

electric field due to pont charge is,

E'=k*q/x^2 ---(1)

E'=(9*10^9)*(7*10^-9)/(10*10^-2)^2

E'=6300 N/C

hence,

part a answer E is smaller than the E'

d)

here,

E=k*q/(x*sqrt(x^2+y^2))

E=(k*q/(x^2))*(1+(y/x)^-2

E=(k*q/(x^2))*(1-y^2/2x^2) ---(2)

from equation no (1) and (2)

y^2/2*x^2=1%

y^2/2*x^2=1/100

y^2/2*x^2=0.01

0.025^2/(2*x^2)=1/100

===>x=0.177 m

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