marout (ssm979) – homework 06 – Turner – (56725)
1
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001
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.
Q
encl
= 0
2.
Q
encl
= 2
ǫ
0
E a
2
3.
Q
encl
= 3
ǫ
0
E a
2
4.
Q
encl
=
ǫ
0
E a
2
correct
5.
Q
encl
=
1
2
ǫ
0
E a
2
6.
insufficient information
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.
The top receives
Φ
top
=
-
E a
2
(inward is negative) and the bottom
Φ
bottom
= 2
E a
2
.
The total electric flux is
Φ
E
=
-
E a
2
+ 2
E a
2
=
E a
2
.
Using Gauss’s Law, the charge inside the
box is
Q
encl
=
ǫ
0
Φ
E
=
ǫ
0
E a
2
.
002
10.0 points
A closed surface with dimensions
a
=
b
=
0
.
385 m and
c
= 0
.
693 m is located as in
the figure.
The electric field throughout the
region is nonuniform and given by
vector
E
= (
α
+
β x
2
)ˆ
ı
where
x
is in meters,
α
= 3 N
/
C, and
β
= 2 N
/
(C m
2
).
E
y
x
z
a
c
b
a
What is the magnitude of the net charge
enclosed by the surface?
Correct answer: 2
.
6612
×
10
-
12
C.
Explanation:
Let :
a
=
b
= 0
.
385 m
,
c
= 0
.
693 m
,
α
= 3 N
/
C
,
and
β
= 2 N
/
(C m
2
)
.
The electric field throughout the region is
directed along the
x
-axis and the direction of
d
vector
A
is perpendicular to its surface. Therefore,
vector
E
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- Fall '10
- TURNER,J
- Charge, Electric charge, Qencl
-
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