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(f) For
r
≥
b
we have
2
total
/4
Eq
r
ε
0
=π
or
3
2
0
.
3
ba
E
r
ρ
3
−
=
Thus, for
r
= 3.00
b
= 6.00
a
, the electric field is
33
9
3
21
2
2
2
00
(2.00 )
7
(1.84 10
C/m )(0.100 m)
7
1.35 N/C.
3
(6.00 )
3
36
3(8.85 10
C /N m )
36
aa
a
E
a
ρρ
εε
−
−
−×
⎛⎞
==
=
=
⎜⎟
×⋅
⎝⎠
50. The field is zero for 0
≤
r
≤
a
as a result of Eq. 2316. Thus,
(a)
E
= 0 at
r
= 0,
(b)
E
= 0 at
r
=
a
/2.00, and
(c)
E
= 0 at
r
=
a
.
For
a
r
b
the enclosed charge
q
enc
(for
a
r
b
) is related to the volume by
q
ra
enc
=−
F
H
G
I
K
J
ρ
ππ
4
3
4
3
.
Therefore, the electric field is
E
q
rr
r
a
r
−
F
H
G
I
K
J
=
−
1
44
4
3
4
0
2
0
2
0
2
πε
ρ
πε
ρ
ε
enc
for
a
≤
r
≤
b
.
(d) For
r
=1.50
a
, we have
9
3
2
2
2
(1.50 )
2.375
(1.84 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Charge

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