terry (ect328) – oldfinal 01 – Turner – (59130)
1
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001
10.0 points
47 cm
47 cm
90
◦
4
μ
C
−
9
μ
C
9
μ
C
What is the magnitude of the electrostatic
force
bardbl
vector
F
bardbl
on the top charge?
The
Coulomb
constant
is
8
.
9875
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 2
.
07138 N.
Explanation:
Let :
Q
= 9
μ
C
,
q
= 4
μ
C
,
and
L
= 47 cm
.
L
θ
q
−
Q
Q
bardbl
vector
F
bardbl
=
k
e

q
1
 
q
2

r
2
By symmetry and the fact that force on charge
q
by +
Q
is repulsive and by
−
Q
is attractive,
F
y
= 0
.
The
x
component of the forces on
q
by
Q
and
−
Q
are equal in magnitude and direction.
Note:
cos 45
◦
=
√
2
2
. Hence,
bardbl
vector
F
bardbl
= 2
k
e
q Q
L
2
√
2
2
= 2 (8
.
9875
×
10
9
N
·
m
2
/
C
2
)
×
(4
μ
C) (9
μ
C)
(47 cm)
2
√
2
2
=
2
.
07138 N
.
keywords:
002
10.0 points
A circular arc has a uniform linear charge
density of 9 nC
/
m.
The
value
of
the
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
54
◦
2
.
5 m
x
y
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Correct answer: 29
.
3779 N
/
C.
Explanation:
Let :
λ
= 9 nC
/
m = 9
×
10
−
9
C
/
m
,
Δ
θ
= 54
◦
,
and
r
= 2
.
5 m
.
θ
is defined as the angle in the counter
clockwise direction from the positive
x
axis
as shown in the figure below.
27
◦
27
◦
r
vector
E
θ
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terry (ect328) – oldfinal 01 – Turner – (59130)
2
First,
position
the
arc
symmetrically
around the
y
axis, centered at the origin. By
symmetry (in this rotated configuration) the
field in the
x
direction cancels due to charge
from opposites sides of the
y
axis, so
E
x
= 0
.
For a continuous linear charge distribution,
vector
E
=
k
e
integraldisplay
dq
r
2
ˆ
r
In polar coordinates
dq
=
λ
(
r dθ
)
,
where
λ
is the linear charge density.
The
positive
y
axis is
θ
= 90
◦
, so the
y
component
of the electric field is given by
dE
y
=
dE
sin
θ .
Note:
By symmetry, each half of the arc
about the
y
axis contributes equally to the
electric field at the origin. Hence, we may just
consider the righthalf of the arc (beginning
on the positive
y
axis and extending towards
the positive
x
axis) and multiply the answer
by 2.
Note:
The upper angular limit
θ
= 90
◦
.
The lower angular limit
θ
= 90
◦
−
27
◦
= 63
◦
,
is the angle from the positive
x
axis to the
righthand end of the arc.
E
=
−
2
k
e
parenleftBigg
λ
r
integraldisplay
90
◦
63
◦
sin
θ dθ
parenrightBigg
ˆ
=
−
2
k
e
λ
r
[cos (63
◦
)
−
cos (90
◦
)] ˆ
.
Since
k
e
λ
r
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(9
×
10
−
9
C
/
m)
(2
.
5 m)
= 32
.
3552 N
/
C
,
E
=
−
2 (32
.
3552 N
/
C)
×
[(0
.
45399)
−
(0)] ˆ
=
−
29
.
3779 N
/
C ˆ
bardbl
vector
E
bardbl
=
29
.
3779 N
/
C
.
Alternate Solution:
Just solve for
bardbl
vector
E
bardbl
in
a straight forward manner, positioning the
beginning of the arc on the positive
x
axis (as
shown in the original figure in the question).
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 Fall '09
 Turner
 Correct Answer, Electric charge, Terry

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