(a) What is the total charge? (b) Find the electric field on the x axis at x = 6
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Question
(a) What is the total charge?
(b) Find the electric field on the x axis at x = 6 m.
(c) Find the electric field on the x axis at x = 10 m.
(d) Find the electric field on the x axis at x = 290 m.
(e) Find the field at x = 290 m, using the approximation that the charge is a point charge at the origin.
Compare your result with that for the exact calculation in part (d).
(field for the point charge divided by the field for the line charge)
A uniform line charge of linear charge density = 2.6 nC/m extends from x = 0 to x = 5 m. What is the total charge? Find the electric field on the x axis at x = 6 m. Find the electric field on the x axis at x = 10 m. Find the electric field on the x axis at x = 290 m. Find the field at x = 290 m, using the approximation that the charge is a point charge at the origin. Compare your result with that for the exact calculation in part (d). (field for the point charge divided by the field for the line charge)Explanation / Answer
lets take a small length element dx, q is the total charge, l is total length, a is distance from nearer end.
dq=(q/l)dx
dE=dq/{4??o(a+x)^2 }
substitute for dq:
dE = qdx / {4??ol(a+x)^2}
E = q / 4??ol?_0^l?(a+x) ^(-2)dx (integrating giving proper limits upper limit=l, lower limit= 0)
E=q / {4??o(a+l)}
solve for total charge q=2.6 nC/m*5m=13 nC,
E at 6m is l = 5m, a = 6 - 5 = 1m
E =19.49 N/C
E at 10m is l=5, a= 5 m
E = 2.6*10613/4pi 8.85*10^-12*10 = 11.6 N/C
E at 250 m = 13nC/4pi 8.85*10^-12 *290 = 0.403
(e) use columb's law directly, taking q=13 nC, r= 290m,
E=q / {4??o(r)^2} = 13nC/4pi*8.85*10^-12 * 290*290 = 1.39 *10^-3 N/C
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