In: Physics
An infinite straight wire has a linear charge density of 10 nC/m. What is the electric
field 0.50 meters away?
Electric Field of a Uniformly Charged Wire
Consider a long straight wire which carries the uniform charge
per unit length . We expect the electric field
generated by such a charge distribution to possess cylindrical
symmetry. We also expect the field to point radially (in a
cylindrical sense) away from the wire (assuming that the wire is
positively charged).
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Let us draw a cylindrical gaussian surface, co-axial with the
wire, of radius and length
--see Fig. 11. The above symmetry
arguments imply that the electric field generated by the wire is
everywhere perpendicular to the curved surface of the cylinder.
Thus, according to Gauss' law,
![]() |
(70) |
where is the electric field-strength a
perpendicular distance
from the wire. Here, the
left-hand side represents the electric flux through the gaussian
surface. Note that there is no contribution from the two flat ends
of the cylinder, since the field is parallel to the surface there.
The right-hand side represents the total charge enclosed by the
cylinder, divided by
. It follows that
![]() |
(71) |
The field points radially (in a cylindrical sense) away from the
wire if , and radially towards the wire
if
.
E = 10*10-9/2*3.14*8.85*10-12*.5
E = 359.854