nanni (arn437) – HW#3 – Erskine – (56905)
1
This printout should have 21 questions.
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beFore answering.
001
10.0 points
±our charges are fxed at the corners oF a
square centered at the origin as Follows:
q
at
(
−
a,
+
a
); 2
q
at (+
a,
+
a
);
−
3
q
at (+
a,
−
a
);
and 6
q
at (
−
a,
−
a
). A fFth charge +
q
with
mass
m
is placed at the origin and released
From rest.
±ind the speed when it is a great distance
From the origin, where the potential energy oF
the fFth charge due to the Four point charges
is negligible.
1.
b
vV
b
=
q
r
2
√
5
k
ma
2.
b
b
=
q
r
3
√
5
k
3.
b
b
=
q
r
6
√
6
k
4.
b
b
=
q
r
2
√
2
k
5.
b
b
=
q
r
3
√
2
k
6.
b
b
=
q
r
6
√
2
k
correct
7.
b
b
=
q
r
6
√
5
k
8.
b
b
=
q
r
6
√
3
k
9.
b
b
=
q
r
3
√
3
k
10.
b
b
=
q
r
3
√
6
k
Explanation:
x
a
+6
q
a
−
3
q
+2
q
+
q
+
q, m
√
2
a
The initial energy oF the charge is
E
i
=
K
i
+
U
i
=
U
i
=
q
p
k q
√
2
a
+
2
k q
√
2
a
+
(
−
3
q
)
k
√
2
a
+
6
k q
√
2
a
P
=
6
k q
2
√
2
a
.
The fnal energy is
E
f
=
1
2
mV
2
.
±rom energy conversation, we have
E
i
=
E
f
6
k q
2
√
2
a
=
1
2
2
V
=
q
r
6
√
2
k
.
002
10.0 points
Two insulating spheres having radii 0
.
2 cm
and 0
.
4 cm, masses 0
.
14 kg and 0
.
55 kg, and
charges
−
4
μ
C and 2
μ
C are released From rest
when their centers are separated by 0
.
9 m
.
How Fast is the smaller sphere moving when
they collide?
Correct answer: 11
.
6441 m
/
s.
Explanation:
Let :
k
e
= 8
.
99
×
10
9
N
·
m
2
/
C
2
,
r
1
= 0
.
2 cm = 0
.
002 m
,
r
2
= 0
.
4 cm = 0
.
004 m
,
m
1
= 0
.
14 kg
,
m
2
= 0
.
55 kg
,
q
1
=
−
4
μ
C =
−
4
×
10
−
6
C
,
q
2
= 2
μ
C = 2
×
10
−
6
C
,
and
d
= 0
.
9 m
.
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View Full Documentnanni (arn437) – HW#3 – Erskine – (56905)
2
By conservation of momentum
0 =
m
1
v
1
−
m
2
v
2
v
2
=
m
1
v
1
m
2
By conservation of energy
(
K
+
U
)
i
= (
K
+
U
)
f
0 +
k
e
q
1
q
2
d
=
1
2
m
1
v
2
1
+
1
2
m
2
v
2
2
+
k
e
q
1
q
2
r
1
+
r
2
k
e
q
1
q
2
d
=
1
2
m
1
v
2
1
+
1
2
m
2
p
m
1
v
1
m
2
P
2
+
k
e
q
1
q
2
r
1
+
r
2
−
k
e
q
1
q
2
p
1
r
1
+
r
2
−
1
d
P
=
m
1
v
2
1
2
m
2
(
m
1
+
m
2
)
.
So
v
2
1
=
−
2
m
2
k
e
q
1
q
2
m
1
(
m
1
+
m
2
)
p
1
r
1
+
r
2
−
1
d
P
=
−
2 (0
.
55 kg)
(
8
.
99
×
10
9
N
·
m
2
/
C
2
)
(0
.
14 kg) (0
.
14 kg + 0
.
55 kg)
×
(
−
4
×
10
−
6
C) (2
×
10
−
6
C)
×
p
1
0
.
002 m + 0
.
004 m
−
1
0
.
9 m
P
= 135
.
584 m
2
/
s
2
.
and
v
1
=
r
135
.
584 m
2
/
s
2
=
11
.
6441 m
/
s
.
003
10.0 points
The work needed to carry a 6 C charge from
point A to point B is 23 J.
Calculate the potential diFerence between
point A and B.
Correct answer: 3
.
83333 V.
Explanation:
Let :
q
= 6 C
and
W
= 23 J
.
The work is
W
=
q V
V
=
W
q
=
23 J
6 C
=
3
.
83333 V
.
004
10.0 points
Assume we are given an electric ±eld set up
by an unknown charge distribution.
U
0
is
the amount of work needed to bring a point
charge of charge
q
0
in from in±nity to a point
P
.
If the charge
q
0
is returned to in±nity, how
much work would it take to bring a new charge
of 4
q
0
from in±nity to point
P
?
1.
8
U
0
2.
U
0
4
3.
U
0
4.
U
0
2
5.
0
6.
2
U
0
7.
4
U
0
correct
8.
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 Fall '08
 gositz
 Charge, Electrostatics, Electric charge, KE, Erskine

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