billing (cab4763) – hw 2 – opyrchal – (121104)
1
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001(part1of2)10.0points
Consider three charges arranged as shown.
−
+
+
4
.
2 cm
3
.
3 cm
3
.
9
μ
C
7
.
8
μ
C
−
3
.
5
μ
C
What is the magnitude of the electric field
strength at a point 2
.
3 cm to the left of the
middle charge?
The value of the Coulomb
constant is 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 1
.
37962
×
10
8
N
/
C.
Explanation:
Let :
q
1
= 7
.
8
μ
C = 7
.
8
×
10
−
6
C
,
q
2
= 3
.
9
μ
C = 3
.
9
×
10
−
6
C
,
q
3
=
−
3
.
5
μ
C =
−
3
.
5
×
10
−
6
C
,
r
1
,
2
= 4
.
2 cm = 0
.
042 m
,
r
2
,
3
= 3
.
3 cm = 0
.
033 m
,
x
= 2
.
3 cm = 0
.
023 m
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
r
1
=
r
1
,
2
−
x
= 0
.
042 m
−
0
.
023 m = 0
.
019 m
r
2
=
x
= 0
.
023 m
r
3
=
r
2
,
3
+
x
= 0
.
033 m + 0
.
023 m = 0
.
056 m
vector
E
net
=
vector
E
1
+
vector
E
2
+
vector
E
3
E
=
k
e
q
r
2
Considering the magnitudes of the electric
fields at a point 2
.
3 cm to the left of the middle
charge,
E
1
=
k
e
q
1
r
2
1
is directed away from the charge since
q
1
is
positive,
E
2
=
k
e
q
2
r
2
2
is directed away from the charge since
q
2
is
positive, and
E
3
=
k
e

q
3

r
2
3
is directed toward the charge since
q
3
is nega
tive. Thus
E
net
=
E
1
−
E
2
+
E
3
=
k
e
parenleftbigg
q
1
r
2
1
−
q
2
r
2
2
+

q
3

r
2
3
parenrightbigg
=
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
bracketleftbigg
(7
.
8
×
10
−
6
C)
(0
.
019 m)
2
−
(3
.
9
×
10
−
6
C)
(0
.
023 m)
2
+
(3
.
5
×
10
−
6
C)
(0
.
056 m)
2
bracketrightbigg
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
[(0
.
0216066 N
/
C)
−
(0
.
0073724 N
/
C)
+(0
.
00111607 N
/
C)]
=
1
.
37962
×
10
8
N
/
C
,
directed along the positive
x
axis.
002(part2of2)10.0points
What is the magnitude of the force on a
−
3
.
5
μ
C charge placed at this point?
Correct answer: 482
.
866 N.
Explanation:
Let :
q
=
−
3
.
5
μ
C =
−
3
.
5
×
10
−
6
C
.
The electric force is
F
electric
=
q E
net
=
−
3
.
5
×
10
−
6
C
×
(1
.
37962
×
10
8
N
/
C)
=
−
482
.
866 N
,
bardbl
vector
F
electric
bardbl
=
482
.
866 N
.
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billing (cab4763) – hw 2 – opyrchal – (121104)
2
F
electric
has a magnitude of 482
.
866 N and is
directed along the negative
x
axis.
003(part1of2)10.0points
Three point charges are placed at the vertices
of an equilateral triangle.
2
.
3 m
60
◦
−
3
.
4 C
−
3
.
4 C
−
3
.
4 C
P
ˆ
ı
ˆ
Find the magnitude of the electric field vec
tor
bardbl
vector
E
bardbl
at
P
.
The value of the Coulomb
constant is 8
.
9875
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 7
.
70195
×
10
9
N
/
C.
Explanation:
Let :
a
= 2
.
3 m
,
q
=
−
3
.
4 C
,
and
k
= 8
.
9875
×
10
9
N
·
m
2
/
C
2
.
a
q
q
q
P
ˆ
ı
ˆ
BasicConcepts:
Electric field.
Electric field vectors due to bottom two
charges cancel out each other. The magnitude
of the field vector due to charge at to top of
the triangle, which gives
bardbl
vector
E
bardbl
=
k q
parenleftBigg
√
3
2
a
parenrightBigg
2
=
4
3
k q
a
2
=
4
3
(8
.
9875
×
10
9
N
·
m
2
/
C
2
) (
−
3
.
4 C)
(2
.
3 m)
2
=
7
.
70195
×
10
9
N
/
C
,
where
h
=
a
cos(30
◦
) =
√
3
2
a
is the height of
the triangle.
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 Spring '10
 Fayngold
 Physics, Charge, Correct Answer, Electric charge, Felectric

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