song (shs546) – Homework 2 – weathers – (22202)
1
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printout
should
have
12
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0points
Four point charges are placed at the four cor
ners of a square, where each side has a length
a
. The upper two charges have identical pos
itive charge, and the two lower charges have
charges of the same magnitude as the first two
but opposite sign. That is,
q
1
=
q
2
=
q
and
q
3
=
q
4
=
−
q
where
q >
0.
q
4
q
2
q
3
q
1
P
j
i
Determine the direction of the electric field
at the center, point P.
1.i
2.
−
i
3.j
4.
0
5.
−
1
√
2
(
i
−
j
)
6.
1
√
2
(
i
−
j
)
7.
1
√
2
(
i
+
j
)
8.
−
jcorrect
9.
−
1
√
2
(
i
+
j
)
Explanation:
The direction is already clear:
all the
x

components cancel, and the lower charges at
tract and the top ones repel, so the answer is
−
j
.
002(part1of2)10.0points
Three positive charges are arranged as shown
at the corners of a rectangle.
+
+
+
0
.
321 m
0
.
963 m
4
.
4 nC
2
.
2 nC
2
.
35 nC
Find the magnitude of the electric field
at the fourth corner of the rectangle.
The
Coulomb constant is 8
.
98755
×
10
9
N m
2
/
C
2
.
Correct answer: 207
.
862 N
/
C.
Explanation:
r
1
r
2
θ
ϕ
Q
2
Q
1
Q
3
E
1
E
2
E
3
Let :
q
1
= 2
.
2 nC = 2
.
2
×
10
−
9
C
,
q
2
= 4
.
4 nC = 4
.
4
×
10
−
9
C
,
q
3
= 2
.
35 nC = 2
.
35
×
10
−
9
C
,
r
2
= 0
.
963 m
,
r
1
= 0
.
321 m
,
and
k
e
= 8
.
98755
×
10
9
N m
2
/
C
2
.
The length of the diagonal is
r
3
=
radicalBig
r
2
1
+
r
2
2
=
radicalBig
(0
.
963 m)
2
+ (0
.
321 m)
2
= 1
.
01509 m
.
The electric field produced by the charge
q
1
is along the positive
y
axis, so
E
1
,x
= 0
E
1
,y
=
E
1
=
k
c
q
1
r
2
1
.
The electric field produced by the charge
q
2
is along the negative
x
axis, so
E
2
,x
=
−
E
2
=
−
k
c
q
2
r
2
2
E
2
,y
= 0
.
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song (shs546) – Homework 2 – weathers – (22202)
2
The electric field produced by the charge
q
3
is upward and to the left along the diagonal of
the rectangle, and is directed away from the
charge since
q
3
is positive, so
E
3
=
k
e
q
3
r
2
3
.
Consider the direction of
E
3
cos
ϕ
=
r
2
r
3
and
sin
ϕ
=
r
1
r
3
Thus
E
3
,x
=
−
k
e
q
3
r
2
3
sin
ϕ
=
−
k
e
q
3
r
2
r
3
3
E
3
,y
=
k
C
q
3
r
2
3
cos
ϕ
=
k
e
q
3
r
1
r
3
3
,
so that
E
net,x
=
E
2
,x
+
E
3
,x
=
−
k
e
q
2
r
2
2
−
k
e
q
3
r
2
r
3
3
=
−
(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
4
.
4
×
10
−
9
C
(0
.
963 m)
2
−
(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
(2
.
35
×
10
−
9
C) (0
.
963 m)
(1
.
01509 m)
3
=
−
62
.
0879 N
/
C
,
and
E
net,y
=
E
1
,y
+
E
3
,y
=
k
e
q
1
r
2
1
+
k
e
q
3
r
1
r
3
3
=
(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
2
.
2
×
10
−
9
C
(0
.
321 m)
2
+
(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
(2
.
35
×
10
−
9
C) (0
.
321 m)
(1
.
01509 m)
3
= 198
.
373 N
/
C
,
so that
E
net
=
radicalBig
E
net,x
2
+
E
net,
4
2
E
net
=
bracketleftbig
(
−
62
.
0879 N
/
C)
2
+(198
.
373 N
/
C)
2
bracketrightbig
1
/
2
=
207
.
862 N
/
C
.
003(part2of2)10.0points
What is the direction of this electric field (as
an angle measured from the positive
x
axis,
with counterclockwise positive)?
Correct answer: 107
.
379
◦
.
Explanation:
tan
θ
=
E
net,y
E
net,x
θ
= tan
−
1
parenleftbigg
E
net,y
E
net,x
parenrightbigg
= tan
−
1
parenleftbigg
198
.
373 N
/
C
−
62
.
0879 N
/
C
parenrightbigg
= 107
.
379
◦
.
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 Spring '00
 Littler
 Charge, Work, Electric charge

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