soto (rrs946) – Homework 2 – Hofmann – (56605)
1
This printout should have 20 questions.
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the next column or page – Fnd all choices
be±ore answering.
001
10.0 points
A rod 18
.
8 cm long is uni±ormly charged and
has a total charge o±

22
.
1
μ
C.
²ind the magnitude o± the electric Feld
along the axis o± the rod at a point 31
.
7287 cm
±rom the center o± the rod. The Coulomb con
stant is 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 2
.
16284
×
10
6
N
/
C.
Explanation:
Let :
ℓ
= 18
.
8 cm = 0
.
188 m
,
Q
=

22
.
1
μ
C =

2
.
21
×
10

5
C
,
r
= 31
.
7287 cm = 0
.
317287 m
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
²or a rod o± length
ℓ
and linear charge density
(charge per unit length)
λ
, the Feld at a dis
tance
d
±rom the end o± the rod along the axis
is
E
=
k
e
i
d
+
ℓ
d
λ
x
2
dx
=
k
e

λ
x
v
v
v
v
d
+
ℓ
d
=
k
e
λ ℓ
d
(
ℓ
+
d
)
,
where
dq
=
λ dx
. The linear charge density
(i± the total charge is
Q
) is
λ
=
Q
ℓ
so that
E
=
k
e
Q
ℓ
ℓ
d
(
ℓ
+
d
)
=
k
e
Q
d
(
ℓ
+
d
)
.
In this problem, we have the ±ollowing situa
tion (the distance
r
±rom the center is given):
d
l
r
r
The distance
d
is
d
=
r

ℓ
2
= 0
.
317287 m

0
.
188 m
2
= 0
.
223287 m
,
and the magnitude o± the electric Feld is
E
=
k
e
Q
d
(
ℓ
+
d
)
=
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×

2
.
21
×
10

5
C

(0
.
223287 m)(0
.
188 m + 0
.
223287 m)
=
2
.
16284
×
10
6
N
/
C
.
The direction must be
toward
the rod, since
the charge distribution is negative (a positive
test charge would be attracted), so the sign
should be positive, according to the conven
tion stated in the problem.
002
10.0 points
A 15
.
1 g piece o± Styro±oam carries a net
charge o±

0
.
5
μ
C and ³oats above the center
o± a very large horizontal sheet o± plastic that
has a uni±orm charge density on its sur±ace.
What is the charge per unit area on the
plastic sheet? The acceleration o± gravity is
9
.
8 m
/
s
2
and the permittivity o± ±ree space is
8
.
85419
×
10

12
C
2
/
N
/
m
2
.
Correct answer:

5
.
24097
μ
C
/
m
2
.
Explanation:
Let :
m
= 15
.
1 g = 0
.
0151 kg
,
q
=

0
.
5
μ
C =

5
×
10

7
C
,
g
= 9
.
8 m
/
s
2
,
and
ǫ
0
= 8
.
85419
×
10

12
C
2
/
N
/
m
2
.
The Feld
E
=
σ
2
ǫ
0
due to a nonconduct
ing inFnite sheet o± charge is the same as that
very close to any plane uni±orm charge distri
bution, where
σ
is the sur±ace charge density
(charge per unit area) o± the plastic sheet.
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 Fall '11
 HOFFMAN
 Charge, Electrostatics, Work, Electric charge

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