aldawsari (ma28683) – HW02 – Betancourt – (16856)
1
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21.521.7
001
10.0 points
A rod 8 cm long is uniformly charged and has
a total charge of

22
.
7
μ
C.
Determine
the
magnitude
of
the
elec
tric
field
along
the
axis
of
the
rod
at
a
point
54
.
6761
cm
from
the
center
of
the
rod.
The
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Correct answer: 6
.
86124
×
10
5
N
/
C.
Explanation:
Let :
ℓ
= 8 cm
,
Q
=

22
.
7
μ
C
,
r
= 54
.
6761 cm
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
For a rod of length
ℓ
and linear charge density
(charge per unit length)
λ
, the field at a dis
tance
d
from the end of the rod along the axis
is
E
=
k
e
integraldisplay
d
+
ℓ
d
λ
x
2
dx
=
k
e

λ
x
vextendsingle
vextendsingle
vextendsingle
vextendsingle
d
+
ℓ
d
=
k
e
λ ℓ
d
(
ℓ
+
d
)
,
where
dq
=
λ dx
.
The linear charge density
(if 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 following situa
tion (the distance
r
from the center is given):
d
l
r
r
The distance
d
is
d
=
r

ℓ
2
= 54
.
6761 cm

8 cm
2
= 0
.
506761 m
,
and the magnitude of the electric field is
E
=
k
e
Q
d
(
ℓ
+
d
)
=
(
8
.
98755
×
10
9
N
·
m
2
/
C
2
)
×

2
.
27
×
10

5
C

(0
.
506761 m)(0
.
08 m + 0
.
506761 m)
=
6
.
86124
×
10
5
N
/
C
.
Now, 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
convention stated in the problem.
keywords:
002
(part 1 of 3) 10.0 points
Consider the setup shown in the figure be
low, where the arc is a semicircle with radius
r
.
The total charge
Q
is negative, and dis
tributed uniformly on the semicircle.
The
charge on a small segment with angle Δ
θ
is
labeled Δ
q
.
x
y


















Δ
θ
θ
r
x
y
I
II
III
IV
B
A
O
Δ
q
is given by
1.
Δ
q
=
Q
Δ
θ
π
correct
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aldawsari (ma28683) – HW02 – Betancourt – (16856)
2
2.
Δ
q
=
Q
π
3.
Δ
q
=
2
Q
Δ
θ
π
4.
None of these
5.
Δ
q
=
Q
2
π
6.
Δ
q
=
Q
7.
Δ
q
=
π Q
8.
Δ
q
= 2
π Q
9.
Δ
q
=
Q
Δ
θ
2
π
10.
Δ
q
=
2
Q
π
Explanation:
The angle of a semicircle is
π
, thus the
charge on a small segment with angle Δ
θ
is
Δ
q
=
Q
Δ
θ
π
.
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 Spring '10
 Gavrine
 Charge, Correct Answer, Electric charge, Betancourt

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