CBSE Class 12-science Answered
conceptual problem
Asked by manu p.prakash | 20 Apr, 2011, 02:31: AM
Expert Answer
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
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.
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
33
00
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
points in the radial direction.
(
)
0
2
/
E
r
?
??
=
?
?
(5) Applying Gausss law gives
, or
0
2
E
Answered by | 03 May, 2011, 08:58: PM
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