howard (cah3459) – oldhomework 09 – Turner – (56705)
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 3) 10.0 points
Consider a long, uniFormly charged, cylindri
cal insulator oF radius
R
and charge density
1
.
1
μ
C
/
m
3
. (The volume oF a cylinder with
radius
r
and length
!
is
V
=
π r
2
!
.)
R
1 cm
What is the electric feld inside the insulator
at a distance 1 cm From the axis (1 cm
<R
)?
Correct answer: 621
.
175 N
/
C.
Explanation:
Given :
ρ
=1
.
1
μ
C
/
m
3
.
1
×
10

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

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

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

12
C
2
/
N
/
m
2
)
=
621
.
175 N
/
C
.
002
(part 2 oF 3) 10.0 points
Determine the absolute value oF the potential
di²erence between
r
1
and
R
, where
r
1
.
(³or
r < R
the electric feld takes the Form
E
=
Cr
, where
C
is positive.)
1.

V

=
C
±
1
r
1

1
R
²
2.

V

=
C
±
1
r
2
1

1
R
2
²
3.

V

=
C
³
R
2

r
2
1
4.

V

=
C
(
R
2

r
2
1
)
5.

V

=
1
6.

V

=
C
(
R

r
1
)
r
1
7.

V

=
C
(
R

r
1
)
8.

V

=
1
2
C
(
R
2

r
2
1
)
correct
9.

V

=
C
r
1
2
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View Full Documenthoward (cah3459) – oldhomework 09 – Turner – (56705)
2
10.

V

=
1
2
C
(
R

r
1
)
r
1
Explanation:
The potential diference 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
=

±
B
A
"
E
·
"
ds
=

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

±
R
r
1
C r dr
=

C
r
2
2
²
²
²
²
R
r
1
=

C
³
R
2
2

r
2
1
2
´
.
The absolute value oF the potential difer
ence is

Δ
V

=
C
³
R
2
2

r
2
1
2
´
=
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
2.
None oF these
3.
V
r
1
>V
R
correct
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
0 and
R > r
1
.
Thus
V
B
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.
004
(part 1 oF 3) 10.0 points
Consider two concentric spherical conducting
shells. The smaller shell has radius
a
=0
.
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 Spring '10
 Turner
 Electrostatics, Electric charge, Howard, E1 E2 E1

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