gilvin (jg47854) – 5. Gauss’s Law 1 – meyers – (21235)
1
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printout
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have
21
questions.
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before answering.
This homework is due Tuesday, February 8,
at midnight Tucson time.
001(part1of3)10.0points
A uniform electric field
vector
E
=
a
ˆ
ı
+
b
ˆ
, inter
sects a surface of area
A
.
What is the flux through this area if the
surface lies in the
yz
plane?
1.
(
a
−
b
)
A
2.
(
b
−
a
)
A
3.
A
radicalbig
a
2
+
b
2
4.
a A
correct
5.
b A
6.
(
a
+
b
)
A
7.
0
Explanation:
BasicConcepts:
Φ =
contintegraldisplay
Surface
vector
E
·
d
vector
A
For a
uniform
electric field and a flat sur
face, this simplifies to Φ =
vector
E
·
vector
A
.
Solution:
The electric flux through a surface
is given by
Φ =
vector
E
·
vector
A ,
where
vector
E
is the electric field and
vector
A
is the vector
which is directed perpendicular to the surface
and has magnitude equal to the area of the
surface.
When the surface is in the
yz
plane, it is
perpendicular to the
x
direction.
Therefore,
the resulting flux is due to the
x
component
of the electric field only.
This can also be
written as
Φ =
vector
E
·
vector
A
= (
a
ˆ
ı
+
b
ˆ
)
·
A
ˆ
ı
=
a A
.
E
E
E
fig 1
fig 2
fig 3
x
y
z
y
y
x
x
z
z
002(part2of3)10.0points
What is the flux if the surface lies in the
xz
plane?
1.
0
2.
(
a
+
b
)
A
3.
(
a
−
b
)
A
4.
(
b
−
a
)
A
5.
b A
correct
6.
A
radicalbig
a
2
+
b
2
7.
a A
Explanation:
When the surface is in the
xz
plane, it is
perpendicular to the
y
direction. Therefore
Φ = (
a
ˆ
ı
+
b
ˆ
)
·
A
ˆ
=
b A
.
003(part3of3)10.0points
What is the flux if the surface lies in the
xy
plane?
1.
A
radicalbig
a
2
+
b
2
2.
(
a
+
b
)
A
3.
b A
4.
a A
5.
0
correct
6.
(
b
−
a
)
A
7.
(
a
−
b
)
A
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gilvin (jg47854) – 5. Gauss’s Law 1 – meyers – (21235)
2
Explanation:
When the surface is in the
xy
plane, it
is perpendicular to the
z
direction.
There
is no component of the electric field in the
z

direction, so there is no component of the elec
tric field perpendicular to the surface.
This
means that the electric flux through the sur
face is zero. Mathematically, this is written
Φ = (
a
ˆ
ı
+
b
ˆ
)
·
A
ˆ
k
=
0
.
004
10.0points
A 34 cm diameter circular loop is rotated
in
a
uniform
electric
field
until
the
po
sition of maximum electric
flux is
found.
The flux in this position is measured to be
6
.
2
×
10
5
N
·
m
2
/
C.
What is the electric field strength?
Correct answer: 6
.
82879
×
10
6
N
/
C.
Explanation:
Let :
r
= 17 cm = 0
.
17 m
and
Φ
max
= 6
.
2
×
10
5
N
·
m
2
/
C
.
Flux is
Φ =
E A
cos
θ .
θ
= 0
◦
for maximum flux, so
Φ
max
=
E A
E
=
Φ
max
π r
2
=
6
.
2
×
10
5
N
·
m
2
/
C
π
(0
.
17 m)
2
=
6
.
82879
×
10
6
N
/
C
.
005
10.0points
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.
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 Spring '11
 doc
 Charge, Magnetism, Work, Electric charge, Surface, Gauss’ Law, µC

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