nguyen (jmn727) – homework 06 – Turner – (59070)
1
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001
10.0 points
A 114 cm diameter loop is rotated in a uniform
electric field until the position of maximum
electric flux is found. The flux in this position
is measured to be 6
.
12
×
10
5
N
·
m
2
/
C.
What is the electric field strength?
Correct answer: 5
.
99587
×
10
5
N
/
C.
Explanation:
Let :
r
= 57 cm = 0
.
57 m
and
Φ = 6
.
12
×
10
5
N
·
m
2
/
C
.
By Gauss’ law,
Φ =
contintegraldisplay
vector
E
·
d
vector
A
The position of maximum electric flux will be
that position in which the plane of the loop is
perpendicular to the electric field;
i.e.
, when
vector
E
·
d
vector
A
=
E dA
. Since the field is constant,
Φ =
E A
=
Eπ r
2
E
=
Φ
π r
2
=
6
.
12
×
10
5
N
·
m
2
/
C
π
(0
.
57 m)
2
=
5
.
99587
×
10
5
N
/
C
.
002
10.0 points
A (7 m by 7 m) square base pyramid with
height of 4
.
45 m is placed in a vertical electric
field of 55
.
1 N
/
C.
7 m
4
.
45 m
55
.
1 N
/
C
Calculate the total electric flux which goes
out through the pyramid’s four slanted sur
faces.
Correct answer: 2699
.
9 N m
2
/
C.
Explanation:
Let :
s
= 7 m
,
h
= 4
.
45 m
,
and
E
= 55
.
1 N
/
C
.
By Gauss’ law,
Φ =
vector
E
·
vector
A
Since there is no charge contained in the pyra
mid, the net flux through the pyramid must
be 0 N/C. Since the field is vertical, the flux
through the base of the pyramid is equal and
opposite to the flux through the four sides.
Thus we calculate the flux through the base
of the pyramid, which is
Φ =
E A
=
E s
2
= (55
.
1 N
/
C) (7 m)
2
=
2699
.
9 N m
2
/
C
.
003
10.0 points
A spherical shell of radius 4
.
3 m is placed
in a uniform electric field with magnitude
3570 N
/
C.
Determine the total electric flux through
the shell.
Correct answer: 0 N
·
m
2
/
C.
Explanation:
Let :
r
= 4
.
3 m
and
E
= 3570 N
/
C
.
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 Spring '08
 Turner
 Work, Correct Answer, Electric charge, total electric flux

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