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C:\User\Teaching\505-506\F10\HW01sol.doc
PHY 505
Classical Electrodynamics I
Fall 2010
Homework 1 with solutions
Problem 1.1
(to be graded of 10 points). Calculate the electric field created by a thin, long, straight
filament, electrically charged with a constant linear density
, using two approaches:
(i) directly from Eq. (8), and
(ii) using the Gauss law.
Solutions
:
(i) From the translational and axial symmetries of the problem, it is clear that
E
(
r
) =
n
E
(
), where
is the distance from the filament.
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Let us select the plane of drawing so that it contains
both the filament and the observation point, and take the line of filament for axis
z
(see Fig. below).
Then
E
(
) may be calculated as
z
z
z
z
z
z
z
dE
dE
dE
E
2
/
1
2
2
cos
,
where
dE
is the magnitude of the elementary contribution to the field, created by a small segment
dz
of
the filament, with electric charge
dz
. According to Eq. (1.3) of the lecture notes,
2
2
0
4
1
z
dz
dE
,
so that the total field
0
2
/
3
2
0
2
/
3
2
2
0
2
1
4
4
d
z
dz
E
.
(*)
(For the last transition, I have used the well-known table integral – see, e.g., MA Eq. (6.5b).)
(ii) Taking a round cylinder of radius
and length
L
, with its axis on the filament, for the
Gaussian volume, we ensure that the electric field
E
is the same, and perpendicular to the volume
boundary on its side walls, while the field flux through cylinder’s “lids” is zero. As a result, Eq. (1.16)
yields
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I regret using the same letter (
) as for the charge density per unit volume (which is not used in this problem),
but both notations are traditional. Let me hope this will not result in any confusion.
z
dz
0
E
d
dE
2
/
1
2
2
)
(
z