This preview shows pages 1–2. Sign up to view the full content.
problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere
of charge and passes through the point where the electric field is to be found. The field is
uniform on the surface, so
2
4
EdA
rE
⋅=
π
³
G
G
v
, where
r
is the radius of the Gaussian surface.
For
r
<
a
, the charge enclosed by the Gaussian surface is
q
1
(
r
/
a
)
3
. Gauss’ law yields
3
2
11
3
00
4
.
4
qq
r
r
E
aa
π
επ
ε
§·
=
¡
=
¨¸¨¸
©¹
(a) For
r
= 0, the above equation implies
E
= 0.
(b) For
r
=
a
/2, we have
92
2
1
5
2
1
32
2
0
(/
2
)
(8.99 10 N m /C )(5.00 10
C)
5.62 10 N/C.
4
2(2.00 10 m)
qa
E
a
πε
−
−
−
×⋅
×
==
=
×
×
(c) For
r
=
a
, we have
2
1
5
1
22
2
0
(8.99 10 N m /C )(5.00 10
C)
0.112 N/C.
4
(2.00 10 m)
q
E
a
−
−
×
=
×
In the case where
a
<
r
<
b
, the charge enclosed by the Gaussian surface is
q
1
, so Gauss’
law leads to
2
2
4.
4
E
r
=
¡
=
(d) For
r
= 1.50
a
, we have
2
1
5
1
2
0
(8.99 10 N m /C )(5.00 10
C)
0.0499 N/C.
4
(1.50 2.00 10 m)
q
E
r
−
−
×
=
××
(e) In the region
b
<
r
<
c
, since the shell is conducting, the electric field is zero. Thus, for
r
= 2.30
a
, we have
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.
This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
 Spring '10
 Rebello,NobelS
 Charge

Click to edit the document details