(a) What is the total charge? nC (b) Find the electric field on the x axis at x
ID: 2125695 • Letter: #
Question
(a) What is the 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 = 12 m.
N/C
(d) Find the electric field on the x axis at x = 270 m.
N/C
(e) Find the field at x = 270 m, using the approximation that the charge is a point charge at the origin.
N/C
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 lambda = 2.6 nC/m extends from x = 0 to x = 5 m.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/{4pie0(a+x)^2 }
substitute for dq:
dE = qdx / {4pi e0 l(a+x)^2}
E = q / 4pi e0 (a+x) ^(-2)dx (integrating giving proper limits upper limit=l, lower limit= 0)
E=q / {4pi e0(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 == 13 *10^-9./ 4*3.14* 8.85*10^-12 * 6 = 19.4 N/C
E at 12m is l=5, a= 5 m
E = 13*10^13/4pi 8.85*10^-12*10 = 6.49 N/C
E at 270 m = 13nC/4pi 8.85*10^-12 *290 = 0.425 N/C
d. EF = L/2 pi eoR = 2.6*10^-9/2* 3.14* 8.85*10^-12 * 270 =0.1732
when compared, 2.45 times
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