aldawsari (ma28683) – HW04 – Betancourt – (16856)
1
This
printout
should
have
23
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
22.422.5,23.1
001
(part 1 of 2) 10.0 points
A cylindrical shell of radius 9 cm and length
255 cm has its charge density uniformly dis
tributed on its surface.
The electric field
intensity at a point 27
.
2 cm radially outward
from its axis (measured from the midpoint of
the shell ) is 40400 N
/
C.
Given:
k
e
= 8
.
99
×
10
9
N
·
m
2
/
C
2
.
What is the net charge on the shell?
Correct answer: 1
.
55848
×
10
−
6
C.
Explanation:
Let :
a
= 0
.
09 m
,
l
= 2
.
55 m
,
E
= 40400 N
/
C
,
and
r
= 0
.
272 m
.
Applying Gauss’ law
contintegraldisplay
E
·
dA
=
Q
ǫ
0
2
π r ℓ E
=
Q
ǫ
0
E
=
Q
2
π ǫ
0
r ℓ
Q
=
E r ℓ
2
k
=
(40400 N
/
C) (0
.
272 m) (2
.
55 m)
2 (8
.
99
×
10
9
N
·
m
2
/
C
2
)
=
1
.
55848
×
10
−
6
C
.
002
(part 2 of 2) 10.0 points
What is the electric field at a point 4
.
99 cm
from the axis?
Correct answer: 0 N
/
C.
Explanation:
E
= 0 inside the shell.
003
(part 1 of 3) 10.0 points
Consider a solid insulating sphere of radius
b
with nonuniform charge density
ρ
=
a r
,
where
a
is a constant.
r
dr
O
b
Find the charge
Q
r
contained within the
radius
r
, when
r < b
as in the figure.
Note:
The volume element
dV
for a spherical shell of
radius
r
and thickness
dr
is equal to 4
π r
2
dr
.)
1.
Q
r
=
r
3
π a
2.
Q
r
= 0
3.
Q
r
=
a r
4
π
4.
Q
r
=
r
4
π a
5.
Q
r
=
π a r
3
6.
Q
r
=
a r
2
π
7.
Q
r
=
a π
r
2
8.
Q
r
=
π a r
2
9.
Q
r
=
a r
3
π
10.
Q
r
=
π a r
4
correct
Explanation:
Basic Concepts:
dQ
=
ρ dV
; Gauss’ law.
Solution:
A charge element is given by
dq
=
ρ dV
= (
a r
)(4
π r
2
dr
)
= 4
π a r
3
dr
For
r < b
, we integrate to find the total charge
within radius
r
:
Q
=
integraldisplay
dq
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
aldawsari (ma28683) – HW04 – Betancourt – (16856)
2
= 4
π a
integraldisplay
r
0
r
′
3
dr
′
= 4
π a
r
4
4
=
π a r
4
.
004
(part 2 of 3) 10.0 points
If
a
= 6
×
10
−
6
C
/
m
4
and
b
= 1 m, find
E
at
r
= 0
.
5 m.
Correct answer: 42352
.
8 N
/
C.
Explanation:
Pick a sphere of radius
r
, concentric with
the first, as the Gaussian surface (since the
problem exhibits spherical symmetry). Over
this surface, the flux is constant due to sym
metry, so Gauss’ Law gives
4
π r
2
E
=
Q
encl
ǫ
0
Since
Q
encl
=
π a r
4
=
π
(
6
×
10
−
6
C
/
m
4
)
(0
.
5 m)
4
= 1
.
1781
×
10
−
6
C
,
we can substitute this into Gauss’ Law to find
E
=
Q
encl
ǫ
0
·
1
4
π r
2
=
1
.
1781
×
10
−
6
C
8
.
85419
×
10
−
12
C
2
/
N
·
m
2
×
1
4
π
(0
.
5 m)
2
=
42352
.
8 N
/
C
at the radius
r
= 0
.
5 m, directed outward.
005
(part 3 of 3) 10.0 points
Find the charge
Q
b
contained within the ra
dius
r
, when
r > b
.
1.
Q
b
= 0
2.
Q
b
=
π a b
2
3.
Q
b
=
π a b
4
correct
4.
Q
b
=
π b
3
a
5.
Q
b
=
a b
2
π
6.
Q
b
=
a b
4
π
7.
Q
b
=
π b
2
a
8.
Q
b
=
π a b
3
9.
Q
b
=
a b
3
π
10.
Q
b
=
π a
b
4
Explanation:
For
r > b
,
Q
=
integraldisplay
dq
= 4
π a
integraldisplay
b
0
r
3
dr
= 4
π a
b
4
4
=
π a b
4
Obviously, the charge enclosed within radius
r
does not depend on
r
when we go outside
the sphere; it is just the charge of the entire
sphere.
Therefore, we could have used the
result
Q
=
π a r
4
from the Part 1 and just
inserted
r
=
b
instead of carrying out the
integration again.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Gavrine
 Electrostatics, SEPTA Regional Rail, Electric charge, Jaguar Racing

Click to edit the document details