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Gauss’s Law
2227
EXECUTE:
(a)
At this point the field of the lefthand sphere is zero and the field of the righthand sphere is toward
the center of that sphere, in the +
x
direction. This point is outside the righthand sphere, a distance
2
rR
=
from its
center.
2
0
1
ˆ
44
Q
π
R
+
E
=i
G
P
.
(b)
This point is inside the lefthand sphere, at
/2
=
, and is outside the righthand sphere, at
3 /2
=
. Both
fields are in the +
x
direction.
32
2
2
2
00
0
1(
/
2
)
1
4
1
1
7
ˆˆ
ˆ
4(
3
2
)
4
2
9
4
1
8
QR
Q
Q
Q
Q
π
RR
π
π
R
⎡⎤
++
⎢⎥
⎣⎦
E
=
i
=
i
G
PP
P
.
(c)
This point is outside both spheres, at a distance
=
from their centers. Both fields are in the +
x
direction.
22
2
1
42
QQ
Q
π
π
R
+
E
=
i
G
.
(d)
This point is outside both spheres, a distance
3
=
from the center of the lefthand sphere and a distance
=
from the center of the righthand sphere. The field of the lefthand sphere is in the +
x
direction and the field
of the righthand sphere is in the
direction
x
−
.
2
0
11
1
8
ˆ
3
)
49
4
9
Q
Q
Q
π
π
π
R
−
−−
E
=
i
=
i
G
P
.
EVALUATE:
At all points on the
x
axis the net field is parallel to the
x
axis.
22.65.
IDENTIFY:
Let
dQ
−
be the electron charge contained within a spherical shell of radius
r
′
and thickness
dr
′
.
Integrate
r
′
from 0 to
r
to find the electron charge within a sphere of radius
r
. Apply Gauss’s law to a sphere of
radius
r
to find the electric field ()
E r
.
SET UP:
The volume of the spherical shell is
2
4
dV
r
dr
π
′
′
=
.
EXECUTE:
(a)
/
2
33
0
()
r
ar
a
r
Qr
Q
ρ
dV
Q
e
r dr
Q
r e
dr
π
aa
′
−
−
′
′
=−
∫∫
∫
.
0
2/
2
0
4
( )
(2
2
2)
[2( /
)
2( /
)
1].
r
ra
r
Qe
Q
e
α
r
r
Q
e
a
α
−
−
−
− − =
+
+
Note if
, ( )
0
rQ
r
→∞
→
; the total net charge of the atom is zero.
(b)
The electric field is radially outward. Gauss’s law gives
2
0
(4
)
Er
=
P
and
0
2
2
(2(
)
2(
) 1)
kQe
a
r
a
r
−
=+
+
.
(c)
The graph of
E
versus
r
is sketched in Figure 22.65. What is plotted is the scaled
E
, equal to
pt charge
/
EE
, versus
scaled
r
, equal to
0
/
.
pt charge
2
kQ
E
r
=
is the field of a point charge.
EVALUATE:
As
0
r
→
, the field approaches that of the positive point charge (the proton). For increasing
r
the
electric field falls to zero more rapidly than the
2
1/
r
dependence for a point charge.
Figure 22.65
22.66.
IDENTIFY:
The charge in a spherical shell of radius
r
and thickness
dr
is
2
()4
dQ
r
r dr
ρπ
=
. Apply Gauss’s law.
SET UP:
Use a Gaussian surface that is a sphere of radius
r
. Let
i
Q
be the charge in the region
/ 2
≤
and let
0
Q
be the charge in the region where
/ 2
R
≤
≤
.
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View Full Document2228
Chapter 22
EXECUTE:
(a)
The total charge is
0
i
QQ Q
=
+
, where
33
4( 2
)
36
i
π
R
απ
R
Q
α
==
and
44
3
23
0
/2
(8
)
(1
6
)
1
1
4(
2)
(
/ )
8
342
4
R
R
RR
απ
R
Q
πα
rrR
d
r
απ
R
⎛⎞
−−
=−
=
−
=
⎜⎟
⎝⎠
∫
. Therefore,
3
15
24
απ
R
Q
=
and
3
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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