Patel, Kinal – Homework 2 – Due: Jan 25 2008, 7:00 pm – Inst: Weathers
1
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printout
should
have
12
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
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.

1
√
2
(
i

j
)
2.
1
√
2
(
i
+
j
)
3.

1
√
2
(
i
+
j
)
4.
0
5.

j correct
6.

i
7. j
8. i
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
.
keywords:
002
(part 1 of 2) 10 points
Three positive charges are arranged as shown
at the corners of a rectangle.
The
Coulomb
constant
is
8
.
98755
×
10
9
N m
2
/
C
2
.
+
+
+
0
.
752 m
2
.
256 m
11
.
9 nC
5
.
95 nC
3
.
73 nC
Find the magnitude of the electric field at
the fourth corner of the rectangle.
Correct answer: 100
.
049 N
/
C.
Explanation:
r
1
r
2
θ
ϕ
Q
2
Q
1
Q
3
E
1
E
2
E
3
Let :
q
1
= 5
.
95 nC = 5
.
95
×
10

9
C
,
q
2
= 11
.
9 nC = 1
.
19
×
10

8
C
,
q
3
= 3
.
73 nC = 3
.
73
×
10

9
C
,
r
2
= 2
.
256 m
,
r
1
= 0
.
752 m
,
and
k
e
= 8
.
98755
×
10
9
N m
2
/
C
2
.
The length of the diagonal is
r
3
=
q
r
2
1
+
r
2
2
=
q
(2
.
256 m)
2
+ (0
.
752 m)
2
= 2
.
37803 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
.
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Patel, Kinal – Homework 2 – Due: Jan 25 2008, 7:00 pm – Inst: Weathers
2
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
.
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
)
×
1
.
19
×
10

8
C
(2
.
256 m)
2

(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
(3
.
73
×
10

9
C) (2
.
256 m)
(2
.
37803 m)
3
=

26
.
6379 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
)
×
5
.
95
×
10

9
C
(0
.
752 m)
2
+
(
8
.
98755
×
10
9
N m
2
/
C
2
)
×
(3
.
73
×
10

9
C) (0
.
752 m)
(2
.
37803 m)
3
= 96
.
4379 N
/
C
,
so that
E
net
=
q
E
net,x
2
+
E
net,
4
2
E
net
=
£
(

26
.
6379 N
/
C)
2
+(96
.
4379 N
/
C)
2
/
1
2
=
100
.
049 N
/
C
.
003
(part 2 of 2) 10 points
What is the direction of this electric field (as
an angle measured from the positive
x
axis,
with counterclockwise positive)?
Correct answer: 105
.
441
◦
.
Explanation:
tan
θ
=
E
net,y
E
net,x
θ
= tan

1
E
net,y
E
net,x
¶
= tan

1
96
.
4379 N
/
C

26
.
6379 N
/
C
¶
= 105
.
441
◦
.
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 Spring '08
 Weathers
 Work, Electric charge, patel

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