segovia (cs39966) – Homework 1 – Spurlock – (14441)
1
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001
10.0points
Calculate the net charge on a substance con-
sisting of a combination of 8
.
6
×
10
13
protons
and 6
.
8
×
10
13
electrons.
Correct answer: 2
.
88
×
10
−
6
C.
Explanation:
Let :
N
p
= 8
.
6
×
10
13
charges
,
q
p
= 1
.
60
×
10
−
19
C
,
N
e
= 6
.
8
×
10
13
charges
,
and
q
e
=
−
1
.
60
×
10
−
19
C
.
Q
net
=
N
p
q
p
+
N
e
q
e
Q
net
= (8
.
6
×
10
13
)(1
.
6
×
10
−
19
C)
+ (6
.
8
×
10
13
)(
−
1
.
6
×
10
−
19
C)
=
2
.
88
×
10
−
6
C
002
10.0points
A particle of mass 41 g and charge 64
μ
C is
released from rest when it is 78 cm from a
second particle of charge
−
17
μ
C.
Determine the magnitude of the initial ac-
celeration of the 41 g particle.
Correct answer: 392
.
008 m
/
s
2
.
Explanation:
Let :
m
= 41 g
,
q
= 64
μ
C = 6
.
4
×
10
−
5
C
,
d
= 78 cm = 0
.
78 m
,
Q
=
−
17
μ
C =
−
1
.
7
×
10
−
5
C
,
and
k
e
= 8
.
9875
×
10
9
.
The force exerted on the particle is
F
=
k
e
|
q
1
| |
q
2
|
r
2
=
m a
bardbl
vectora
bardbl
=
k
e
bardbl
vectorq
bardbl bardbl
vector
Q
bardbl
m d
2
=
k
e
vextendsingle
vextendsingle
6
.
4
×
10
−
5
C
vextendsingle
vextendsingle
vextendsingle
vextendsingle
−
1
.
7
×
10
−
5
C
vextendsingle
vextendsingle
(0
.
041 kg) (0
.
78 m
2
)
=
392
.
008 m
/
s
2
.
003
10.0points
Three charges are arranged in the (
x, y
) plane
(as shown in the figure below, where the scale
is in meters).
−
5 nC
−
4 nC
9 nC
y
(
m
)
0
1
2
3
4
5
6
7
8
9 10
x
(
m
)
0
1
2
3
4
5
6
7
8
9
10
What is the magnitude of the resulting force
on the
−
5 nC charge at the origin?
Correct answer: 12
.
8898 nN.
Explanation:
Let :
q
o
=
−
5
×
10
−
9
C
,
(
x
o
, y
o
) = (0 m
,
0 m)
,
q
a
=
−
4
×
10
−
9
C
,
(
x
a
, y
a
) = (4 m
,
0 m)
,
q
b
= 9
×
10
−
9
C
,
and
(
x
b
, y
b
) = (0 m
,
8 m)
.
Coulomb’s Law for
q
o
and
q
a
is
bardbl
vector
F
oa
bardbl
=
k
e
q
o
q
a
radicalbig
(
x
a
−
x
o
)
2
+ (
y
a
−
y
o
)
2
=
k
e
q
o
q
a
radicalbig
x
2
oa
+
y
2
oa
= 8
.
98755
×
10
9
N C
2
/
m
2
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segovia (cs39966) – Homework 1 – Spurlock – (14441)
2
×
(
−
5
×
10
−
9
C) (
−
4
×
10
−
9
C)
radicalbig
(4 m)
2
+ (0 m)
2
= 1
.
12344
×
10
−
8
N
.
F
oa
x
=
−
k
e
q
o
q
a
r
2
oa
cos
θ
oa
=
−
k
e
q
o
q
a
r
2
oa
x
oa
r
oa
=
−
(8
.
98755
×
10
9
N C
2
/
m
2
)
×
(
−
5
×
10
−
9
C) (
−
4
×
10
−
9
C)
(4 m)
2
×
4 m
4 m
=
−
1
.
12344
×
10
−
8
N
.
F
oa
y
=
−
k
e
q
o
q
a
r
2
oa
sin
θ
oa
=
−
k
e
q
o
q
oa
r
2
oa
y
oa
r
oa
=
−
(8
.
98755
×
10
9
N C
2
/
m
2
)
×
(
−
5
×
10
−
9
C) (
−
4
×
10
−
9
C)
(4 m)
2
×
0 m
4 m
= 0 N
.
Coulomb’s Law for
q
o
and
q
b
is
bardbl
vector
F
ob
bardbl
=
k
e
q
o
q
b
radicalbig
(
x
b
−
x
o
)
2
+ (
y
b
−
y
o
)
2
=
k
e
q
o
q
b
radicalBig
x
2
ob
+
y
2
ob
= 8
.
98755
×
10
9
N C
2
/
m
2
×
(
−
5
×
10
−
9
C) (9
×
10
−
9
C)
radicalbig
(0 m)
2
+ (8 m)
2
=
−
6
.
31937
×
10
−
9
N
.
F
ob
x
=
−
k
e
q
o
q
a
r
2
oa
cos
θ
ob
=
−
k
e
q
o
q
b
r
2
ob
x
ob
r
ob
=
−
(8
.
98755
×
10
9
N C
2
/
m
2
)
×
(
−
5
×
10
−
9
C) (9
×
10
−
9
C)
(8 m)
2
×
0 m
8 m
= 0 N
.
F
ob
y
=
−
k
e
q
o
q
a
r
2
oa
sin
θ
ob
=
−
k
e
q
o
q
b
r
2
ob
y
ob
r
ob
=
−
(8
.
98755
×
10
9
N C
2
/
m
2
)
×
(
−
5
×
10
−
9
C) (9
×
10
−
9
C)
(8 m)
2
×
8 m
8 m
= 6
.
31937
×
10
−
9
N
.

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- Fall '11
- HOFFMAN
- Charge, Work, Correct Answer, Electric charge, Spurlock, Segovia
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