homework 02 – KIM, JI – Due: Jan 27 2008, 4:00 am
1
Question 1, chap 24, sect 1.
part 1 of 1
10 points
An electric field of magnitude 16000 N
/
C
and directed upward (perpendicular to the
Earth’s surface) exists on a day when a thun
derstorm is brewing.
A truck that can be
approximated as a rectangle 7
.
1 m by 3
.
3 m
is traveling along a road that is inclined 16
◦
relative to the ground.
Determine the electric flux through the bot
tom of the truck.
Correct answer: 360358 N
·
m
2
/
C (tolerance
±
1 %).
Explanation:
Let :
E
= 16000 N
/
C
,
ℓ
= 7
.
1 m
,
w
= 3
.
3 m
,
and
θ
= 16
◦
.
By Gauss’ law,
Φ =
vector
E
·
vector
A .
The flux through the bottom of the car is
Φ =
E A
cos
θ
=
E ℓ w
cos
θ
= (16000 N
/
C) (7
.
1 m) (3
.
3 m) cos 16
◦
=
360358 N
·
m
2
/
C
.
Question 2, chap 24, sect 2.
part 1 of 1
10 points
A cubic box of side
a
, oriented as shown,
contains an unknown charge.
The vertically
directed electric field has a uniform magni
tude
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
=
1
2
ǫ
0
E a
2
2.
Q
encl
= 3
E
ǫ
0
a
2
3.
Q
encl
= 2
ǫ
0
E a
2
4.
Q
encl
= 6
ǫ
0
E a
2
5.
Q
encl
= 3
ǫ
0
E a
2
6.
Q
encl
=
E
ǫ
0
a
2
7.
insufficient information
8.
Q
encl
= 0
9.
Q
encl
= 2
E
ǫ
0
a
2
10.
Q
encl
=
ǫ
0
E a
2
correct
Explanation:
By convention, electric flux through a sur
face
S
is 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 passes
through the vertical sides. The top receives
Φ
top
=
−
E a
2
(inward is negative) and the bottom
Φ
bottom
= 2
E a
2
,
so the total electric flux is
Φ
E
=
−
E a
2
+ 2
E a
2
=
E a
2
.
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homework 02 – KIM, JI – Due: Jan 27 2008, 4:00 am
2
Using Gauss’ Law, the charge inside the box
is
Q
encl
=
ǫ
0
Φ
E
=
ǫ
0
E a
2
.
Question 3, chap 24, sect 2.
part 1 of 1
10 points
A charge of 17
.
2
μ
C is at the geometric
center of a cube.
What is the electric flux through one of
the faces?
The permittivity of free space is
8
.
85419
×
10
−
12
C
2
/
N
·
m
2
.
Correct answer: 323764 N
·
m
2
/
C (tolerance
±
1 %).
Explanation:
Let :
q
= 17
.
2
μ
C = 1
.
72
×
10
−
5
C
and
ǫ
0
= 8
.
85419
×
10
−
12
C
2
/
N
·
m
2
.
By Gauss’ law,
Φ =
contintegraldisplay
vector
E
·
d
vector
A =
q
ǫ
0
.
The total flux through the cube is given by
Φ
tot
=
q
ǫ
0
=
1
.
72
×
10
−
5
C
8
.
85419
×
10
−
12
C
2
/
N
·
m
2
= 1
.
94258
×
10
6
N
·
m
2
/
C
,
so the flux through one side of the cube is
Φ =
1
6
Φ
tot
=
323764 N
·
m
2
/
C
.
Question 4, chap 24, sect 2.
part 1 of 1
10 points
A point charge of 8
.
5
μ
C is located at the
center of a uniform ring having linear charge
density 14
.
2
μ
C
/
m and radius 2
.
85 m.
q
R
a
λ
Find the total electric flux through a sphere
centered at the point charge and having radius
R < a
as shown.
The premittivity of free
space is 8
.
85419
×
10
−
12
C
2
/
N m
2
.
Correct answer:
959998
N m
2
/
C (tolerance
±
1 %).
Explanation:
Let :
q
= 8
.
5
μ
C = 8
.
5
×
10
−
6
C
,
λ
= 14
.
2
μ
C
/
m = 1
.
42
×
10
−
5
C
/
m
,
ǫ
0
= 8
.
85419
×
10
−
12
C
2
/
N m
2
,
and
a
= 2
.
85 m
.
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 Spring '08
 Turner
 Physics, Charge, Electrostatics, Work, Electric charge

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