51. (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
=
³³³
³
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=
¨¸
©¹
or
31
2
3
3
1
5
(14.1 10
C/m )(0.0560 m)
7.78 10
C.
s
QR
πρ
π
−−
==×
=
×
(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:
enc
2
0
1
4
q
E
r
ε
=
π
where
q
enc
is given by an integral similar to that worked in part (a):
4
2
enc
0
44.
4
r
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Rebello,NobelS
 Charge, Trigraph, Electric charge, Fundamental physics concepts, qenc

Click to edit the document details