36
Gauss’s Law
P24.28
σ
=×
F
H
G
I
K
J
−−
860 10
100
6
2
2
..
Ccm
cm
m
Cm
22
ej
E
=
∈
=
×
×
−
−
2
2 8 85 10
486 10
0
2
12
9
.
.
.
N C away from the wall
The field is essentially uniform as long as the distance from the center of the wall to the field point is
much less than the dimensions of the wall.
P24.29
If
ρ
is positive, the field must be radially outward. Choose as the
gaussian surface a cylinder of length
L
and radius
r
, contained inside
the charged rod. Its volume is
π
rL
2
and it encloses charge
ρπ
2
.
Because the charge distribution is long, no electric flux passes
through the circular end caps;
EA
⋅=
°
=
dE
d
A
cos
.
90 0
0 . The curved
surface has
°
d
A
cos0 , and
E
must be the same strength
everywhere over the curved surface.
FIG. P24.29
Gauss’s law,
∈
z
d
q
0
,
becomes
Ed
A
Curved
Surface
z
=
∈
2
0
.
Now the lateral surface area of the cylinder is 2
rL
:
Er
L
2
2
0
bg
=
∈
.
Thus,
E
=
∈
r
2
0
radially away from the cylinder axis .
*P24.30
Let
represent the charge density. For the field inside the sphere at
r
1
5
=
cm we have
qr
11
2
0
1
3
0
4
4
3
πρ
=
∈
=
∈
inside
E
r
1
1
0
3
=
∈
=
∈
=
×−
×
=−
×
−
−
3
3 8 85 10
86 10
005
457 10
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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