Question
Wed April 20, 2011

# conceptual problem

Tue May 03, 2011
The amount of charge enclosed by the Gaussian surface, a cylinder of radius r and length l is
As indicated in figure the gaussian surface consists of three parts:a two ends  plus the curved side wall . The flux through gaussian surface is
+ +
=
we have set  can be seen from the figure and no flux passes throughthe ends since area vector are perpendicular to the electric field which point in the radial direction.

Figure 4.2.7 Gaussian surface for a uniformly charged rod.
(4) As indicated in Figure 4.2.7, the Gaussian surface consists of three parts: a two ends
and
plus the curved side wall
. The flux through the Gaussian surface is
1
1
2
2
3
3
3 3
0 0
E
S
S
S
d
d
d
d
EA
?=
?=
?
+
?
+
?
=++
=
??
??
??
??
EA
EA
E
A
E
A
?
?
?
?
?
1
S
2
S
3
S
?
?
?
?
(
)
3
2
S
E
r
??
(4.2.15)
1
2
where we have set
3
E
E
=. As can be seen from the figure, no flux passes through the
ends since the area vectors
1
dA?and
2
dA?are perpendicular to the electric field which
(
)
0
2
/
E
r
?
??
=
?
?
(5) Applying GaussÂ’s law gives
, or
0
2
E
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