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(d) For
r
=
R
, the electric field is
2
92
2
1
2
3
00
2
1
(8.99 10 N m C ) (14.1 10
C/m )(0.0560 m)
44
2.23 10 N/C.
ss
RR
E
R
πρ
π
πε
−
−
==
=
×
⋅
×
=×
55. (a) We integrate the volume charge density over the volume and require the result be
equal to the total charge:
2
0
4 .
R
dx dy dz
dr r
Q
ρρ
=
π=
∫∫∫
∫
Substituting the expression
ρ
=
s
r
/
R
, with
s
= 14.1 pC/m
3
, and performing the integration
leads to
4
4
4
s
R
Q
R
⎛⎞
π
=
⎜⎟
⎝⎠
or
31
2
3
3
1
5
(14.1 10
C/m )(0.0560 m)
7.78 10
C.
s
QR
−−
==×
=
×
(b) At
r
= 0, the electric field is zero (
E
= 0) since the enclosed charge is zero.
At a certain point within the sphere, at some distance
r
from the center, the field (see Eq.
238 through Eq. 2310) is given by Gauss’ law:
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 Spring '08
 Any
 Physics, Charge

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