Reilly (mar3978) – homework 05 – Turner – (56725)
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001
(part 1 of 2) 10.0 points
A charged mass on the end of a light string
is attached to a point on a uniformly charged
vertical sheet of infinite extent.
The acceleration of gravity is 9
.
8 m
/
s
2
and
the
permittivity
of
free
space
is
8
.
854
×
10
−
12
C
2
/
N
·
m
2
.
40
.
6 cm
θ
ˆ
ı
ˆ
0
.
1
μ
C
1 g
Areal charge density
0
.
24
μ
C
/
m
2
Find the angle
θ
the thread makes with the
vertically charge sheet.
Correct answer: 7
.
87394
◦
.
Explanation:
Let :
g
= 9
.
8 m
/
s
2
,
ǫ
0
= 8
.
854
×
10
−
12
C
2
/
N
·
m
2
,
m
= 1 g = 0
.
001 kg
,
σ
= 0
.
24
μ
C
/
m
2
= 2
.
4
×
10
−
7
C
/
m
2
,
q
= 0
.
1
μ
C = 1
×
10
−
7
C
,
and
L
= 40
.
6 cm = 0
.
406 m
.
The length
L
of the string is superfluous.
Let the tension in the string be denoted by
T
. The electric field due to the infinite sheet
is constant in the
x
direction and is
vector
E
=
σ
2
ǫ
0
ˆ
ı .
In the ˆ
ı
and ˆ
directions, force equilibrium
tells us
T
sin
θ
=
q
·
σ
2
ǫ
0
T
cos
θ
=
m g
tan
θ
=
T
sin
θ
T
cos
θ
=
q σ
2
m g ǫ
0
θ
= arctan
parenleftbigg
q σ
2
m g ǫ
0
parenrightbigg
= arctan
bracketleftbigg
(1
×
10
−
7
C) (2
.
4
×
10
−
7
C
/
m
2
)
2 (0
.
001 kg) (9
.
8 m
/
s
2
)
ǫ
0
bracketrightbigg
=
7
.
87394
◦
.
002
(part 2 of 2) 10.0 points
What value would
σ
in order for he angle 84
◦
?
Correct answer: 16
.
5112
μ
C
/
m
2
.
Explanation:
From the previous part, we know
σ
=
2
m g
tan
θ
q
ǫ
0
=
2 (0
.
001 kg)
(
9
.
8 m
/
s
2
)
tan(84
◦
)
1
×
10
−
7
C
·
(8
.
854
×
10
−
12
C
2
/
N
·
m
2
)
10
6
μ
C
1 C
=
16
.
5112
μ
C
/
m
2
.
003
10.0 points
Two
large,
parallel,
insulating
plates
are
charged uniformly with the same positive
areal charge density +
σ
, which is the charge
per unit area. The permittivity of free space
ǫ
0
=
1
4
π k
e
.
The magnitude of the resultant electric field
E
(where “outside” stands for above and be
low the two plates) is
1.
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 Spring '10
 TURNER,J
 Correct Answer, Electric charge, Reilly

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