Version PREVIEW – HW 2 – feng – (58230)
1
This printout should have 26 questions.
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before answering.
Uniformly Charged Bent Wire 01
001
10.0 points
A uniformly charged insulating rod of length
13
.
2 cm is bent into the shape of a semicircle
as in the Fgure.
The value of the Coulomb constant is
8
.
98755
×
10
9
N m
2
/
C
2
.
1
3
.
2
c
m
−
7
.
48
μ
C
O
If the rod has a total charge of
−
7
.
48
μ
C,
Fnd the horizontal component of the electric
Feld at
O
, the center of the semicircle. DeFne
right as positive.
Correct answer:
−
2
.
42424
×
10
7
N
/
C.
Explanation:
Let :
L
= 13
.
2 cm = 0
.
132 m
and
q
=
−
7
.
48
μ
C =
−
7
.
48
×
10
−
6
C
.
Call the length of the rod
L
and its charge
q
. Due to symmetry
E
y
=
i
dE
y
= 0
and
E
x
=
i
dE
sin
θ
=
k
e
i
dq
sin
θ
r
2
,
where
dq
=
λ dx
=
λ r dθ
, so that
b
y
x
θ
E
x
=
−
k
e
λ
r
i
3
π/
2
π/
2
cos
θ dθ
=
−
k
e
λ
r
(sin
θ
)
v
v
v
v
3
π/
2
π/
2
= 2
k
e
λ
r
,
where
λ
=
q
L
and
r
=
L
π
.
Therefore,
E
x
=
2
k
e
q π
L
2
=
2 (8
.
98755
×
10
9
N m
2
/
C
2
)
(0
.
132 m)
2
×
(
−
7
.
48
×
10
−
6
C)
π
=
−
2
.
42424
×
10
7
N
/
C
.
Since the rod has negative charge, the Feld is
pointing to the left (towards the charge dis
tribution). A positive test charge at O would
feel an attractive force from the semicircle,
pointing to the left.
Charged Arc 03
002
(part 1 of 2) 10.0 points
A uniformly charged circular arc AB of radius
R
is shown in the Fgure. It covers a quarter
of a circle and it is located in the second
quadrant.
The total charge on the arc is
Q >
0.
The value of the Coulomb constant is
8
.
99
×
10
9
N
·
m
2
/
C
2
.
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View Full DocumentVersion PREVIEW – HW 2 – feng – (58230)
2
x
y
+
+
+
+
+
+
+
+
+
Δ
θ
θ
R
x
y
I
II
III
IV
B
A
O
Δ
s
≡
R
Δ
θ
The direction of the electric Feld vector
v
E
at the origin, due to the charge distribution,
is
1.
along the negative
y
axis
2.
along the negative
x
axis
3.
along the positive
y
axis
4.
in quadrant III
5.
in quadrant II
6.
in quadrant I
7.
in quadrant IV
correct
8.
along the positive
x
axis
Explanation:
Basic Concept:
Electric Felds generated
by charge distributions.
Solution:
The electric Feld for a positive
charge is directed away from it. In this case,
the electric Feld generated by each Δ
q
will be
directed into quadrant IV, so the total electric
Feld will be in the same quadrant.
003
(part 2 of 2) 10.0 points
±ind
E
x
, the
x
component of the electric Feld
at the origin due to the full arc length for the
case, where
Q
= 0
.
7
μ
C and
R
= 1
.
59 m.
Correct answer: 1584
.
26 N
/
C.
Explanation:
Let :
Q
= 0
.
7
μ
C
and
R
= 1
.
59 m
.
E
x
can be found by integrating the con
tributions of all the Δ
q
’s in the arc.
Us
ing Δ
q
=
Q
π/
2
Δ
θ
and then adding the x
component of the Feld from each charge ele
ment Δ
q
at angle
θ
, we have
i
π/
2
0
k
e
Q
R
2
2
π
cos
θ dθ
=
2
k
e
Q
π R
2
=
2
k
e
(0
.
7
μ
C)
π
(1
.
59 m)
2
=
1584
.
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 Spring '08
 Turner
 Physics, Charge, Electrostatics, Electric charge

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