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CHAPTER
23
The Electric Field II: Continuous Charge Distributions
1*
∙
A uniform line charge of linear charge density
λ
= 3.5 nC/m extends from
x
= 0 to
x
= 5 m. (
a
) What is the
total charge? Find the electric field on the
x
axis at (
b
)
x
= 6 m, (
c
)
x
= 9 m, and (
d
)
x
= 250 m. (
e
) Find the field
at
x
= 250 m, using the approximation that the charge is a point charge at the origin, and compare your result
with that for the exact calculation in part (
d
).
(
a
)
Q
=
L
(
b
), (
c
), (
d
)
E
x
(
x
0
) =
kQ
/[
x
0
(
x
0
–
L
)], Equ. 235
(
e
)
E
x
≅
kQ
/
x
2
Q
= (3.5
×
10
–9
×
5) C = 17.5 nC
E
x
(6) = 26.2 N/C;
E
x
(9) = 4.37 N/C;
E
x
(250) = 2.57
×
10
–3
N/C
E
x
(250) = 2.52
×
10
–3
N/C, within 2% of (
d
)
2
∙
Two infinite vertical planes of charge are parallel to each other and are separated by a distance
d
= 4 m.
Find the electric field to the left of the planes, to the right of the planes, and between the planes (
a
) when each
plane has a uniform surface charge density
σ
= +3
µ
C/m
2
and (
b
) when the left plane has a uniform surface
charge density
= +3
C/m
2
and that of the right plane is
= –3
C/m
2
. Draw the electric field lines for each
case.
(
a
)
E
= 4
π
k
= 3.39
×
10
5
N/C
The field pattern is shown in the adjacent figure.
The field between the plates is zero.
(
b
) Again,
E
= 3.39
×
10
5
N/C.
The field pattern is shown in the adjacent figure.
The field is confined to the region between the
two plates and is zero elsewhere.
3
∙
A 2.75
C charge is uniformly distributed on a ring of radius 8.5 cm. Find the electric field on the axis at
(
a
) 1.2 cm, (
b
) 3.6 cm, and (
c
) 4.0 m from the center of the ring. (
d
) Find the field at 4.0 m using the
approximation that the ring is a point charge at the origin, and compare your results with that for part (
c
).
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View Full DocumentChapter 23
The Electric Field II: Continuous Charge Distributions
(
a
) Use Equ. 2310
(
b
), (
c
) Proceed as in (
a
)
(
d
)
E
x
~
kQ
/
x
2
E
x
=
N/C
10
_
4.69
=
N/C
)
(0.085)
+
((0.012)
0.012
_
10
_2.75_
10
8.99_
5
3/2
2
2
6
9
•
E
x
(0.036) = 1.13
×
10
6
N/C;
E
x
(4) = 1.54
×
10
3
N/C
E
x
= 1.55
×
10
3
N/C; this is slightly greater than (
c
)
because the point charge is nearer
x
= 4 m than the ring.
4
∙
A disk of radius 2.5 cm carries a uniform surface charge density of 3.6
µ
C/m
2
. Using reasonable
approximations, find the electric field on the axis at distances of (
a
) 0.01 cm, (
b
) 0.04 cm, (
c
) 5 m, and (
d
) 5
cm.
For
x
<<
r
, the disk appears like an infinite plane. For
x
>>
r
, the ring charge may be approximated by a point
charge.
(
a
), (
b
) Use Equ. 2312
a
(
c
)
E
x
=
kQ
/
x
2
=
k
π
r
2
σ
/
x
2
(d)
E
x
=
k
r
2
/
x
2
; this is not a good approximation
since
x
= 2
r
is not much greater than
r
.
E
x
= 2.03
×
10
5
N/C
E
x
= 2.54 N/C
E
x
= 2.54
×
10
4
N/C
5*
∙
For the disk charge of Problem 4, calculate exactly the electric field on the axis at distances of (
a
) 0.04 cm
and (
b
) 5 m, and compare your results with those for parts (
b
) and (
c
) of Problem 4.
(
a
) Use Equ. 2311;
r
= 2.5 cm,
= 3.6
C/m
2
(
b
) Proceed as in (
a
)
E
x
= 2.00
×
10
5
N/C
E
x
= 2.54 N/C
For
x
= 0.04 cm, the exact value of
E
x
is only 1.5% smaller than the approximate value obtained in the
preceding problem. For
x
= 5 m, the exact and approximate values agree within less than 1%.
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 Spring '08
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