howard (cah3459) – homework 04 – Turner – (56705)
1
This printout should have 10 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 4) 10.0 points
Consider a disk oF radius 3
.
3 cm with a uni
Formly distributed charge oF +4
.
1
μ
C.
Compute the magnitude oF the electric feld
at a point on the axis and 2
.
4 mm From the
center. The value oF the Coulomb constant is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 6
.
2766
×
10
7
N
/
C.
Explanation:
Let :
R
=3
.
3 cm = 0
.
033 m
,
k
e
=8
.
98755
×
10
9
N
·
m
2
/
C
2
,
Q
=4
.
1
μ
C = 4
.
1
×
10

6
C
,
and
x
=2
.
4 mm = 0
.
0024 m
.
The surFace charge density is
σ
=
Q
π R
2
=
4
.
1
×
10

6
C
π
(0
.
033 m)
2
=0
.
00119841 C
/
m
2
.
The feld at the distance
x
along the axis oF
a disk with radius
R
is
E
π k
e
σ
±
1

x
√
x
2
+
R
2
²
,
Since
1

x
√
x
2
+
R
2
=1

0
.
0024 m
³
(0
.
0024 m)
2
+ (0
.
033 m)
2
.
927464
,
E
π
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(0
.
00119841 C
/
m
2
)
×
(0
.
927464)
=6
.
2766
×
10
7
N
/
C
so
±
±
E
±
=
6
.
2766
×
10
7
N
/
C
,
002
(part 2 oF 4) 10.0 points
Compute the feld From the nearfeld ap
proximation
x
²
R.
Correct answer: 6
.
76749
×
10
7
N
/
C.
Explanation:
x
²
R
, so the second term in the parenthe
sis can be neglected and
E
approx
e
σ
π
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(0
.
00119841 C
/
m
2
)
=
6
.
76749
×
10
7
N
/
C
close to the disk.
003
(part 3 oF 4) 10.0 points
Compute the electric feld at a point on the
axis and 37 cm From the center oF the disk.
Correct answer: 2
.
67572
×
10
5
N
/
C.
Explanation:
Let :
x
= 37 cm
.
E
e
σ
π
(8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×
(0
.
00119841 C
/
m
2
)
×
´
1

0
.
37 m
³
(0
.
37 m)
2
+ (0
.
033 m)
2
µ
=
2
.
67572
×
10
5
N
/
C
.
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 Fall '08
 Turner
 Electrostatics, Electric charge, Fundamental physics concepts, KE, Etot

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