A uniform line charge that has a linear charge density ? = 4.7 nC/m is on the x

Question

A uniform line charge that has a linear charge density ? = 4.7 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 = 8.0 m.
____________ N/C

(d) Find the electric field on the x axis at x = 290 m.
___________ N/C

(e) Estimate the electric field at x = 290 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.

Explanation / Answer

(a) Linear charge density =Q /L Given = 4.7*10^-9 C/m L = 5m Charge Q = *L = (4.7*10^-9 C/m)(5.0m)=23.5*10^-9C   = 23.5nC (b)electric field at a point P which is at a distance a from last end of the rod electric field E= kQ /(a)(L+a) given a = 6 m - 5m = 1m E =  (9*10^9 N,m^2/C^2)(23.5nC)/(1m)(1m +5m) E=35.25 N/C (c) Given a =8m -5m = 3m electric field E= kQ /(a)(L+a) E =  (9*10^9 N,m^2/C^2)(23.5nC)/(3m)(3m +5m) E=8.8125 N/C (d)a =290m-5m=285m E =  (9*10^9 N,m^2/C^2)(23.5nC)/(285m)(285m +5m) E_d=0.0025589 N/C (e)The electric field due to a point charge E= kq /r^2 Here r = 290 m - 2.5 m   = 287.5 m E = (9*10^9 N,m^2/C^2)(23.5nC)/(287.5m)^2 E_e=0.00255879N/C (f) The ration E_e /E_d = (0.00255879N/C) /0.0025589 N/C =0.99 electric field E= kQ /(a)(L+a) E =  (9*10^9 N,m^2/C^2)(23.5nC)/(3m)(3m +5m) E=8.8125 N/C (d)a =290m-5m=285m E =  (9*10^9 N,m^2/C^2)(23.5nC)/(285m)(285m +5m) E_d=0.0025589 N/C (e)The electric field due to a point charge E= kq /r^2 Here r = 290 m - 2.5 m   = 287.5 m E = (9*10^9 N,m^2/C^2)(23.5nC)/(287.5m)^2 E_e=0.00255879N/C (f) The ration E_e /E_d = (0.00255879N/C) /0.0025589 N/C =0.99 E =  (9*10^9 N,m^2/C^2)(23.5nC)/(285m)(285m +5m) E_d=0.0025589 N/C (e)The electric field due to a point charge E= kq /r^2 Here r = 290 m - 2.5 m   = 287.5 m E = (9*10^9 N,m^2/C^2)(23.5nC)/(287.5m)^2 E_e=0.00255879N/C (f) The ration E_e /E_d = (0.00255879N/C) /0.0025589 N/C =0.99

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