terry (ect328) – oldhomework 02 – Turner – (59130)
1
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printout
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13
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before answering.
001
(part 1 of 2) 10.0 points
Three charges are arranged as shown in the
figure.
The
Coulomb
cosnstant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
x
y
3
.
7 nC
4
.
3 m
−
0
.
9 nC
0
.
6 nC
−
1
.
9 m
Find the magnitude of the electrostatic
force on the charge at the origin.
Correct answer: 2
.
10413 nN.
Explanation:
Let :
(
x
0
, y
0
) = (0 m
,
0 m)
,
q
0
=
−
0
.
9 nC
,
(
x
1
, y
1
) = (4
.
3 m
,
0 m)
,
q
1
= 3
.
7 nC
,
(
x
2
, y
2
) = (0 m
,
−
1
.
9 m)
,
and
q
2
= 0
.
6 nC
.
x
y
q
1
x
1
q
0
q
2
y
2
F
θ
The force
vector
F
10
(in the
hatwider
10 direction,
−
ˆ
ı
) be
tween charge
−
9
×
10
−
10
C and 3
.
7
×
10
−
9
C
is
vector
F
10
= +
k
e
q
0
q
1
x
2
1
(
−
ˆ
ı
)
= (
−
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(
−
0
.
9 nC) (3
.
7 nC)
(4
.
3 m)
2
ˆ
ı
= (1
.
61863
×
10
−
9
N)ˆ
ı .
The plus sign means the force
vector
F
10
is towards
the positive
x
axis. The force
vector
F
20
(in the
hatwider
20
direction, +ˆ
) between charge
−
9
×
10
−
10
C
and 6
×
10
−
10
C is
vector
F
20
= +
k
e
q
0
q
2
y
2
2
(+ˆ
)
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(
−
0
.
9 nC) (0
.
6 nC)
(
−
1
.
9 m)
2
ˆ
= (
−
1
.
3444
×
10
−
9
N)ˆ
.
The minus sign means the force
vector
F
20
is
towards the negative
y
axis.
The magni
tude of the total force exerted on the charge
−
9
×
10
−
10
C is equal to
bardbl
vector
F
bardbl
=
radicalBig
F
2
10
+
F
2
20
=
bracketleftBig
(1
.
61863
×
10
−
9
N)
2
+ (
−
1
.
3444
×
10
−
9
N)
2
bracketrightBig
1
2
= 2
.
10413
×
10
−
9
N
=
2
.
10413 N
.
002
(part 2 of 2) 10.0 points
What is the angle
θ
between the electrostatic
force on the charge at the origin and the pos
itive
x
axis?
Answer in degrees as an angle
between
−
180
◦
and 180
◦
measured from the
positive
x
axis, with counterclockwise posi
tive.
Correct answer:
−
39
.
7122
◦
.
Explanation:
The angle between
vector
F
and the negative
y

axis is
θ
= arctan
parenleftbigg
−
1
.
3444
×
10
−
9
N
1
.
61863
×
10
−
9
N
parenrightbigg
=
−
39
.
7122
◦
.
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terry (ect328) – oldhomework 02 – Turner – (59130)
2
keywords:
003
(part 1 of 2) 10.0 points
A charge
Q
is distributed uniformly along the
x
axis from
x
1
to
x
2
.
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
Which of the following integrals is correct
for the magnitude of the electric field at
x
0
on
the
x
axis? Assume that
x
0
> x
2
> x
1
and
k
e
=
1
4
πǫ
0
.
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 Fall '09
 Turner
 Electric charge, Terry, KE

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