Nguyen, Thanh – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers
1
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printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
At distance
r
from a point charge
q
, the elec
tric potential is 277 V and the magnitude of
the electric field is 233 N
/
C.
Determine the value of
q
.
Correct answer: 3
.
66406
×
10

8
C.
Explanation:
Let :
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
V
= 277 V
,
and
e
= 233 N
/
C
.
E
=
k
e
q
r
2
and
V
=
k
e
q
r
,
so that
V
E
=
r
.
The potential is
V
=
k
e
q
r
=
k
e
q
V
E
=
k
e
q E
V
q
=
V
2
k
e
E
=
(277 V)
2
(8
.
98755
×
10
9
N
·
m
2
/
C
2
) (233 N
/
C)
=
3
.
66406
×
10

8
C
.
keywords:
002
(part 1 of 1) 10 points
Four
identical
particles
each
have
charge
0
.
2
μ
C and mass 0
.
01 kg. They are released
from rest at the vertices of a square of side
0
.
47 m.
How fast is each charge moving when their
distances from the center of the square dou
bles?
Correct answer: 0
.
321765 m
/
s.
Explanation:
Let :
m
= 0
.
01 kg
,
q
= 0
.
2
μ
C
,
and
L
= 0
.
47 m
.
Each charge moves off on its diagonal line,
and by symmetry each has the same speed.
The initial potential energy of the system of
charges is
U
i
=
k
e
q
2
4
L
+
2
L
√
2
¶
.
The final potential energy is obtained by dou
bling the distances.
By conservation of energy
(
K
+
U
)
i
= (
K
+
U
)
f
4
k
e
q
2
L
+
2
k
e
q
2
L
√
2
=
4
2
m v
2
+
4
k
e
q
2
2
L
+
2
k
e
q
2
2
L
√
2
k
e
q
2
L
2 +
1
√
2
¶
= 2
m v
2
v
=
s
k
e
q
2
m L
1 +
1
2
√
2
¶
Since
k
e
q
2
m L
=
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
(0
.
01 kg)(0
.
47 m)
×
(2
×
10

7
C)
2
= 0
.
0764898 m
2
/
s
2
,
v
=
s
(0
.
0764898 m
2
/
s
2
)
1 +
1
2
√
2
¶
=
0
.
321765 m
/
s
.
keywords:
003
(part 1 of 4) 10 points
The electric potential over a certain region of
space is given by
V
=
a x
2
y

b x z

c y
2
,
where
a
= 8 V
/
m
3
,
b
= 5 V
/
m
2
, and
c
=
8 V
/
m
2
.
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Nguyen, Thanh – Homework 6 – Due: Feb 8 2008, 7:00 pm – Inst: Weathers
2
Find the electric potential at the point
(
x, y, z
) = (8 m
,
1 m
,
1 m).
Correct answer: 464 V.
Explanation:
Let :
a
= 8 V
/
m
3
,
b
= 5 V
/
m
2
,
c
= 8 V
/
m
2
,
and
(
x, y, z
) = (8 m
,
1 m
,
1 m)
.
The electric potential is
V
=
a x
2
y

b x z

c y
2
=
(
8 V
/
m
3
)
(8 m)
2
(1 m)

(
5 V
/
m
2
)
(8 m) (1 m)

(
8 V
/
m
2
)
(1 m)
2
=
464 V
.
004
(part 2 of 4) 10 points
Find the
x
component of the electric field at
the same point.
Correct answer:

123 V
/
m.
Explanation:
The electric field is the gradient of the po
tential, so
E
x
=

∂ V
∂x
=

2
a x y
+
b z
=

2
(
8 V
/
m
3
)
(8 m) (1 m)
+
(
5 V
/
m
2
)
(1 m)
=

123 V
/
m
.
005
(part 3 of 4) 10 points
Find the
y
component of the electric field at
the same point.
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 Spring '00
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 Electrostatics, Work, Electric charge, Fundamental physics concepts, charge density, KE

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