oliver (jmo673) – HW03 – Tsoi – (59090)
1
This
printout
should
have
24
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 4) 10.0 points
For the configuration shown in the figure, sup
pose that
a
= 7 cm,
b
= 21 cm, and
c
= 28 cm.
Furthermore, suppose that the electric field at
a point 10 cm from the center is measured to
be 4400 N
/
C radially inward while the elec
tric field at a point 40 cm from the center is
260 N
/
C radially outward.
a
b
c
Insulator
Conductor
Find the charge on the insulating sphere.
Correct answer:

4
.
89566
×
10
−
9
C.
Explanation:
Let :
a
= 7 cm
,
b
= 21 cm
,
c
= 28 cm
,
r
1
= 10 cm = 0
.
1 m
,
and
E
1
= 4400 N
/
C
,
Basic Concept:
E
= 0 inside conductors,
Gauss’ Law:
contintegraldisplay
vector
E
·
d
vector
A
=
Q
ǫ
0
Solution:
By Gauss’ Law,
E
2
(
4
π r
2
1
)
=
q
ǫ
0
,
where
q
is the charge contained in the insulat
ing sphere, so
q
=
E
2
(
4
π r
2
1
)
ǫ
0
= (4400 N
/
C)
bracketleftbig
4
π
(0
.
1 m)
2
bracketrightbig
×
(8
.
85419
×
10
−
12
C
2
/
N
·
m
2
)
=

4
.
89566
×
10
−
9
C
,
where E is negative because it points radially
inward
.
002
(part 2 of 4) 10.0 points
Find
the
net
charge
on
the
conducting
sphere.
Correct answer: 9
.
52428
×
10
−
9
C.
Explanation:
Let :
r
2
= 40 cm
= 0
.
4 m
and
E
2
= 260 N
/
C
.
Similarly, by Gauss’ Law, the electric field at
r
=
r
2
, where
r
2
> c
, will be related to the to
tal charge enclosed by a concentric Gaussian
sphere of radius
r
2
. If we call the total charge
on the spherical shell
Q
, Gauss’ Law is
E
2
(4
π r
2
2
) =
Q
+
q
ǫ
0
,
from which we find
Q
=
E
2
(
4
π r
2
2
)
ǫ
0

q
= (260 N
/
C)
bracketleftbig
4
π
(0
.
4 m)
2
bracketrightbig
×
(8
.
85419
×
10
−
12
C
2
/
N
·
m
2
)

(

4
.
89566
×
10
−
9
C)
=
9
.
52428
×
10
−
9
C
.
003
(part 3 of 4) 10.0 points
Find the total charge on the inner surface of
the hollow conducting sphere.
Correct answer: 4
.
89566
×
10
−
9
C.
Explanation:
Since
E
= 0 inside a conductor, the to
tal charge contained in a concentric Gaussian
sphere of radius
r
in
, where
b < r
in
< c
, must
be zero.
From this we know that the total
charge on the inner surface of the spherical
conducting shell must be
Q
in
=

q
=
4
.
89566
×
10
−
9
C
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
oliver (jmo673) – HW03 – Tsoi – (59090)
2
004
(part 4 of 4) 10.0 points
Find the total charge on the outer surface of
the hollow conducting sphere.
Correct answer: 4
.
62862
×
10
−
9
C.
Explanation:
To find the total charge on the outer surface
of the conductor,
Q
out
, we note that
Q
out
+
Q
in
=
Q
. Then
Q
out
is
Q
out
=
Q

Q
in
= 9
.
52428
×
10
−
9
C

4
.
89566
×
10
−
9
C
=
4
.
62862
×
10
−
9
C
.
005
(part 1 of 2) 10.0 points
An electron moving parallel to the
x
axis has
an initial speed of 1
×
10
6
m
/
s at the origin.
Its speed is reduced to 6
×
10
5
m
/
s at the point
x
P
, 2 cm away from the origin. The mass of
the electron is 9
.
10939
×
10
−
31
kg and the
charge of the electron is

1
.
60218
×
10
−
19
C.
Calculate the absolute value of the poten
tial difference between this point and the ori
gin.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '09
 TSOI
 Physics, Electrostatics, Electric charge, KE

Click to edit the document details