maldonado (om2892) – Homework
3
2010
UPII – Stone – (13750)
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
A uniformly charged insulating rod of length
16
.
3 cm is bent into the shape of a semicircle
as in the figure.
The
value
of
the
Coulomb
constant
is
8
.
98755
×
10
9
N m
2
/
C
2
.
1
6
.
3
c
m
−
6
.
62
μ
C
O
If the rod has a total charge of
−
6
.
62
μ
C,
find the horizontal component of the electric
field at
O
, the center of the semicircle. Define
right as positive.
Correct answer:
−
1
.
40703
×
10
7
N
/
C.
Explanation:
Let :
L
= 16
.
3 cm = 0
.
163 m
and
q
=
−
6
.
62
μ
C =
−
6
.
62
×
10

6
C
.
Call the length of the rod
L
and its charge
q
. Due to symmetry
E
y
=
integraldisplay
dE
y
= 0
and
E
x
=
integraldisplay
dE
sin
θ
=
k
e
integraldisplay
dq
sin
θ
r
2
,
where
dq
=
λ dx
=
λ r dθ
, so that
y
x
θ
E
x
=
−
k
e
λ
r
integraldisplay
3
π/
2
π/
2
cos
θ dθ
=
−
k
e
λ
r
(sin
θ
)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
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
.
163 m)
2
×
(
−
6
.
62
×
10

6
C)
π
=
−
1
.
40703
×
10
7
N
/
C
.
Since the rod has negative charge, the field 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.
002
(part 1 of 5) 2.0 points
Consider a cylindrical distribution [dark gray]
extending from
r
= 0 to
r
= 9
.
7 cm, of charge
ρ
=
r
a
0
, where
a
0
= 30 cm
·
m
3
/μ
C.
This
cylindrical charge distribution [dark gray] is
surrounded by a dielectric shell [light gray]
whose inner radius is 17 cm and outer radius
is 2
.
6 cm as shown in the figure.
L
2
.
6 cm
17 cm
9
.
7 cm
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maldonado (om2892) – Homework
3
2010
UPII – Stone – (13750)
2
What is the electric field at 3
.
66 cm?
As
sume the length
L
is very long compared to
the diameter of the dielectric shell.
Neglect
edge effects.
Correct answer: 168
.
101 V
/
m.
Explanation:
Let :
r
= 3
.
66 cm
.
Gauss’ law
contintegraldisplay
vector
E dvectora
=
Q
enc
ǫ
0
=
1
ǫ
0
integraldisplay
r
0
ρ
(
r
)
dV
where
dV
=
d
(
π r
2
L
)
= 2
π r L dr .
In our case, Gauss’ law gives
2
π r L E
=
1
ǫ
0
a
0
integraldisplay
r
0
r
·
2
π r L dr,
E
=
r
2
3
ǫ
0
a
0
=
1
8
.
85419
×
10

12
C
2
/
N
·
m
2
×
(3
.
66 cm)
2
3 (30 cm
·
m
3
/μ
C)
×
1 m
100 cm
×
1 C
10
6
μ
C
=
168
.
101 V
/
m
.
003
(part 2 of 5) 2.0 points
What is the electric field at 12
.
7 cm?
Correct answer: 901
.
821 V
/
m.
Explanation:
Let :
r
= 12
.
7 cm
.
Now the integration in the RHS of Gauss’
law goes up to the boundary of our charge
distribution
2
π r E
=
1
ǫ
0
a
0
integraldisplay
R
0
r
·
2
π r dr,
giving
E
=
R
3
3
ǫ
0
a
0
r
=
1
8
.
85419
×
10

12
C
2
/
N
·
m
2
×
(9
.
7 cm)
3
3 (30 cm
·
m
3
/μ
C) (12
.
7 cm)
×
1 m
100 cm
×
1 C
10
6
μ
C
=
901
.
821 V
/
m
.
004
(part 3 of 5) 2.0 points
What is the electric field at 20
.
9 cm?
Correct answer: 307
.
863 V
/
m.
Explanation:
Let :
r
= 20
.
9 cm
and
κ
= 1
.
78
.
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 Summer '10
 Stone
 Charge, Electrostatics, Work, Correct Answer, Electric charge, maldonado

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