sanne (as42476) – Homework 02 – Yao – (59110)
1
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printout
should
have
23
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
The electron gun in a television tube is used
to accelerate electrons (mass of 9
.
10939
×
10

31
kg and charge of
−
1
.
60218
×
10

19
C)
from rest to 2
×
10
7
m
/
s within a distance of
6
.
7 cm.
What electric field is required?
Correct answer: 16972 N
/
C.
Explanation:
Let :
m
e
= 9
.
10939
×
10

31
kg
,
q
e
= 1
.
60218
×
10

19
C
,
v
= 2
×
10
7
m
/
s
,
and
d
= 6
.
7 cm
.
The magnitude of the force is
F
=
q
e
E
=
m
e
a
a
=
q
e
E
m
e
The final velocity is
v
2
f
=
v
2
i
+ 2
a d
= 2
a d
since
v
i
= 0, so
v
2
=
2
d q
e
E
m
e
E
=
v
2
m
e
2
d q
e
=
parenleftBig
2
×
10
7
m
/
s
2
parenrightBig
(
9
.
10939
×
10

31
kg
)
2 (6
.
7 cm) (1
.
60218
×
10

19
C)
=
16972 N
/
C
.
002
10.0 points
A 124 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 3
.
23
×
10
5
N
·
m
2
/
C.
What is the electric field strength?
Correct answer: 2
.
67466
×
10
5
N
/
C.
Explanation:
Let :
r
= 62 cm = 0
.
62 m
and
Φ = 3
.
23
×
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
=
3
.
23
×
10
5
N
·
m
2
/
C
π
(0
.
62 m)
2
=
2
.
67466
×
10
5
N
/
C
.
003
10.0 points
A (4
.
13 m by 4
.
13 m) square base pyramid
with height of 4
.
92 m is placed in a vertical
electric field of 38
.
8 N
/
C.
4
.
13 m
4
.
92 m
38
.
8 N
/
C
Calculate the total electric flux which goes
out through the pyramid’s four slanted sur
faces.
Correct answer: 661
.
808 N m
2
/
C.
Explanation:
Let :
s
= 4
.
13 m
,
h
= 4
.
92 m
,
and
E
= 38
.
8 N
/
C
.
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sanne (as42476) – Homework 02 – Yao – (59110)
2
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
= (38
.
8 N
/
C) (4
.
13 m)
2
=
661
.
808 N m
2
/
C
.
004
10.0 points
A cubic box of side
a
, oriented as shown, con
tains an unknown charge.
The vertically di
rected electric field has a uniform magnitude
E
at the top surface and 2
E
at the bottom
surface.
a
E
2
E
How much charge
Q
is inside the box?
1.
insufficient information
2.
Q
encl
= 0
3.
Q
encl
=
ǫ
0
E a
2
correct
4.
Q
encl
= 3
ǫ
0
E a
2
5.
Q
encl
= 2
ǫ
0
E a
2
6.
Q
encl
=
1
2
ǫ
0
E a
2
Explanation:
Electric flux through a surface
S
is, by con
vention, positive for electric field lines going
out of
the surface
S
and negative for lines
going in.
Here the surface is a cube and no flux goes
through the vertical sides.
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 Spring '08
 Turner
 Physics, Charge, Electrostatics, Work, Electric charge

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