HW 1 - segovia(cs39966 Homework 1 Spurlock(14441 This...

segovia (cs39966) – Homework 1 – Spurlock – (14441) 1
This print-out should have 34 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
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

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 .