? = E
E A
= (a?? + b?) E A?
gilvin (jg47854)  5. Gauss’s Law 1  meyers  (21235)
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This homework is due Tuesday, February 8,
at midnight Tucson time.
001 (part 1 of 3) 10.0 points
A uniform electric field
E
= a ˆı + b ĵ, 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
a
2
+ b
2
4. a A correct
5. b A
6. (a + b) A
7.0
Explanation:
Basic Concepts:
∮
Φ =
E · dA
Surface
For a uniform electric field and a flat sur
face, this simplifies to Φ = E
· A
.
Solution: The electric flux through a surface
is given by
Φ = E
· A ,
where E is the electric field and 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 xdirection.
Therefore,
the resulting flux is due to the xcomponent
of the electric field only.
This can also be
written as
ı =
a A
.
x
x
x
z
z
z
fig 1
fig 2
fig 3
002 (part 2 of 3) 10.0 points
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
a
2
+ b
2
7. a A
Explanation:
When the surface is in the xz plane, it is
perpendicular to the ydirection. Therefore
Φ = (aˆı + bĵ) · Aĵ =
b A
.
003 (part 3 of 3) 10.0 points
What is the flux if the surface lies in the xy
plane?
√
1. A
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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View Full Documentgilvin (jg47854)  5. Gauss’s Law 1  meyers  (21235)
2
Explanation:
When
the
surface
is
in
the
xy
plane,
it
E
is
perpendicular
to
the
zdirection.
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
a
means that the electric flux through the sur
face is zero. Mathematically, this is written
2 E
Φ = (aˆı + bĵ) · Ak
How much charge Q is inside the box?
=
0
.
1. Q
encl
= 3
ϵ
0
E a
2
004
10.0 points
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
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 Spring '11
 doc
 Electrostatics, Magnetism, Work, Electric charge, Surface, Meyers

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