Kapoor (mk9499) – oldhomework 08 – Turner – (60230)
1
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12
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001
(part 1 of 3) 10.0 points
Consider a long, uniformly charged, cylindri
cal insulator of radius
R
and charge density
1
.
4
μ
C
/
m
3
.
(The volume of a cylinder with
radius
r
and length
ℓ
is
V
=
π r
2
ℓ
.)
R
1
.
7 cm
What is the electric field inside the insulator
at a distance 1
.
7 cm from the axis (1
.
7 cm
<
R
)?
Correct answer: 1344 N
/
C.
Explanation:
Given :
ρ
= 1
.
4
μ
C
/
m
3
= 1
.
4
×
10

6
C
/
m
3
,
r
= 1
.
7 cm = 0
.
017 m
,
and
ǫ
0
= 8
.
85419
×
10

12
C
2
/
N
/
m
2
.
Consider a cylindrical Gaussian surface of
radius
r
and length
ℓ
much less than the
length of the insulator so that the compo
nent of the electric field parallel to the axis is
negligible.
ℓ
r
R
The flux leaving the ends of the Gaussian
cylinder is negligible, and the only contribu
tion to the flux is from the side of the cylinder.
Since the field is perpendicular to this surface,
the flux is
Φ
s
= 2
π r ℓ E ,
and the charge enclosed by the surface is
Q
enc
=
π r
2
ℓ ρ .
Using Gauss’ law,
Φ
s
=
Q
enc
ǫ
0
2
π r ℓ E
=
π r
2
ℓ ρ
ǫ
0
.
Thus
E
=
ρ r
2
ǫ
0
=
(
1
.
4
×
10

6
C
/
m
3
)
(0
.
017 m)
2 (8
.
85419
×
10

12
C
2
/
N
/
m
2
)
=
1344 N
/
C
.
002
(part 2 of 3) 10.0 points
Determine the absolute value of the potential
difference between
r
1
and
R
, where
r
1
< R
.
(For
r < R
the electric field takes the form
E
=
C r
, where
C
is positive.)
1.

V

=
C
(
R

r
1
)
r
1
2.

V

=
C
(
R
2

r
2
1
)
3.

V

=
1
2
C
(
R

r
1
)
r
1
4.

V

=
C
parenleftbigg
1
r
1

1
R
parenrightbigg
5.

V

=
C r
1
6.

V

=
C
parenleftbigg
1
r
2
1

1
R
2
parenrightbigg
7.

V

=
1
2
C
(
R
2

r
2
1
)
correct
8.

V

=
C
radicalBig
R
2

r
2
1
9.

V

=
C
(
R

r
1
)
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Kapoor (mk9499) – oldhomework 08 – Turner – (60230)
2
10.

V

=
C
r
1
2
Explanation:
The potential difference between a point A
inside the cylinder a distance
r
1
from the axis
to a point B a distance
R
from the axis is
Δ
V
=

integraldisplay
B
A
vector
E
·
vector
ds
=

integraldisplay
R
r
1
E dr ,
since
E
is radial.
Δ
V
=

integraldisplay
R
r
1
C r dr
=

C
r
2
2
vextendsingle
vextendsingle
vextendsingle
vextendsingle
R
r
1
=

C
parenleftbigg
R
2
2

r
2
1
2
parenrightbigg
.
The absolute value of the potential differ
ence is

Δ
V

=
C
parenleftbigg
R
2
2

r
2
1
2
parenrightbigg
=
1
2
C
(
R
2

r
2
1
)
.
003
(part 3 of 3) 10.0 points
What is the relationship between the poten
tials
V
r
1
and
V
R
?
1.
V
r
1
> V
R
correct
2.
None of these
3.
V
r
1
=
V
R
4.
V
r
1
< V
R
Explanation:
Since
C >
0 and
R > r
1
, from Part 2,
V
B

V
A
= Δ
V
=

1
2
C
(
R
2

r
2
1
)
<
0
since
C >
0 and
R > r
1
.
Thus
V
B
< V
A
and the potential is higher
at point A where
r
=
r
1
than at point B,
where
r
=
R .
Intuitive Reasoning:
The natural ten
dency for a positive charge is to move from A
to B, so A has a higher potential.
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 Fall '08
 Turner
 Electrostatics, Work, Electric charge, Kapoor

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