benavides (jjb2356) – oldfinal 01 – Turner – (59130)
1
This
printout
should
have
31
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
26 cm
26 cm
90
◦
5
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: 8
.
46096 N.
Explanation:
Let :
Q
= 9
μ
C
,
q
= 5
μ
C
,
and
L
= 26 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
)
×
(5
μ
C) (9
μ
C)
(26 cm)
2
√
2
2
=
8
.
46096 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
.
245
◦
2
.
7 m
x
y
What is the magnitude of the electric field
at the center of the circle along which the arc
lies?
Correct answer: 50
.
5335 N
/
C.
Explanation:
Let :
λ
=
−
9 nC
/
m =
−
9
×
10
−
9
C
/
m
,
Δ
θ
= 245
◦
,
and
r
= 2
.
7 m
.
θ
is defined as the angle in the counter
clockwise direction from the positive
x
axis
as shown in the figure below.
122
.
5
◦
122
.
5
◦
r
vector
E
θ
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
benavides (jjb2356) – 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
◦
−
122
.
5
◦
=
−
32
.
5
◦
, is the angle from the positive
x
axis
to the righthand end of the arc.
E
=
−
2
k
e
parenleftBigg
λ
r
integraldisplay
90
◦
−
32
.
5
◦
sin
θ dθ
parenrightBigg
ˆ
=
−
2
k
e
λ
r
[cos (
−
32
.
5
◦
)
−
cos (90
◦
)] ˆ
.
Since
k
e
λ
r
= (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(
−
9
×
10
−
9
C
/
m)
(2
.
7 m)
=
−
29
.
9585 N
/
C
,
E
=
−
2 (
−
29
.
9585 N
/
C)
×
[(0
.
843391)
−
(0)] ˆ
= 50
.
5335 N
/
C ˆ
bardbl
vector
E
bardbl
=
50
.
5335 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).
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Magnetic Field, Correct Answer, Electric charge

Click to edit the document details