brown (twb493) – Ch16h2 – chiu – (56565)
1
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printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001(part1of2)10.0points
A thin glass rod of length 75 cm is rubbed
all over with wool and acquires a charge of
65 nC, distributed uniformly over its surface.
Calculate the magnitude of the electric field
due to the rod at a location 8 cm from the
midpoint of the rod.
First, use the exact
formula.
Also,
note
that
the
value
of
k
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 19044
.
5 N
/
C.
Explanation:
The exact formula for the magnitude of the
electric field of a uniformly charged rod is
E
ex
=
1
4
πǫ
0
bracketleftBigg
Q
r
radicalbig
r
2
+ (
L/
2)
2
bracketrightBigg
.
So, plugging in the necessary values, we
obtain
E
ex
=
1
4
πǫ
0
bracketleftBigg
65 nC
(8 cm
radicalbig
(8 cm)
2
+ (75 cm
/
2)
2
)
bracketrightBigg
=
19044
.
5 N
/
C
.
Remember to convert to the correct units.
002(part2of2)10.0points
Now find the electric field using the approxi
mate formula.
Correct answer: 19473 N
/
C.
Explanation:
The approximate formula is
E
ap
=
1
4
πǫ
0
2(
Q/L
)
r
.
Plugging in the necessary values, we obtain
E
ap
=
1
4
πǫ
0
2(65 nC
/
75 cm)
8 cm
=
19473 N
/
C
.
Remember to convert to the correct units.
003(part1of3)10.0points
A small, thin hollow spherical glass shell
of radius
R
carries a uniformly distributed
positive charge +
Q
. Below it is a horizontal
permanent dipole with charges +
q
and
−
q
separated by a distance
s
.
The dipole is fixed and not free to rotate.
The distance from the center of the glass shell
to the center of the dipole is
L
.
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
L
≫
s
s
+
q
−
q
Calculate the
x
component of the electric
field at the center of the shell.
Your answer should only involve the quan
tities
R
,
k
,
q
,
s
and
L
and possibly constant
numbers like 1,
−
5 and
π
(which should be
represented as the letters ”pi”).
Correct answer:
k
∗
q
∗
s/
(
L
∗
L
∗
L
).
Explanation:
The charge on the shell is all on its sur
face.
Therefore any gaussian surface drawn
totally within the shell will contain no charge.
By Gauss’ Law, the electric flux through the
gaussian surface, and therefore also the elec
tric field, must be zero.
Now, let’s consider the field due to the
dipole.
Both of the charges are equidistant
from the center of the shell.
Furthermore,
since the field points away from the positive
charge and toward the negative charge we
conclude that the
x
component due to both
charges is positive.
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brown (twb493) – Ch16h2 – chiu – (56565)
2
Now, remembering that the field due to a
point charge is
vector
E
=
k
qvectorr
r
3
, we conclude that the
x
component due to each point charge is
k
qx
r
3
,
where
x
is the
x
component of the distance to
the point charge and
r
is the magnitude of the
distance to the point charge. Since
L
≫
s
, we
approximate the total distance to each dipole
charge as
L
rather than
radicalBigg
L
2
+
parenleftbigg
1
2
s
parenrightbigg
2
.
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 Fall '08
 Turner
 Physics, Shell, Electric charge

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