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nguyen (jmn727) – homework 04 – Turner – (59070)
1
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beFore answering.
001
(part 1 oF 3) 10.0 points
A rod oF length
ℓ
with uniForm charge per unit
length
λ
, where
λ >
0, is placed a distance
d
From the origin along the
x
axis. A similar
rod with the same charge is placed along the
y
axis as in the fgure. Consider only a small
segment oF length Δ
x
(it is oF infnitesimal
length) on the horizontal rod, which is placed
along the
x
axis. The segment is at a distance
x
From the origin.
x
y
d
ℓ
λ
O
d
ℓ
λ
Determine the electric feld Δ
v
E
at the ori
gin due to this segment.
1.
Δ
v
E
=
k
e
λ
Δ
x
x
ˆ
ı
2.
Δ
v
E
=
−
k
e
λ
Δ
x
x
ˆ
ı
3.
Δ
v
E
=
k
e
λ
Δ
x
2
x
ˆ
ı
4.
Δ
v
E
=
−
k
e
λ
Δ
x
2
x
ˆ
ı
5.
Δ
v
E
=
−
k
e
λ
Δ
x
x
2
ˆ
ı
correct
6.
Δ
v
E
=
k
e
λ
Δ
x
x
2
ˆ
ı
Explanation:
±rom the Fact that
λ >
0 and the diagram,
the feld Δ
E
at
O
due to the segment on the
x
axis is in the negative
x
direction, and its
magnitude is given by
Δ
E
=
k
e
Δ
q
x
2
=
k
e
λ
Δ
x
x
2
.
We can thus express the electric feld as
Δ
v
E
=
−
Δ
E
ˆ
ı .
002
(part 2 oF 3) 10.0 points
Determine the electric feld
v
E
rod 1
at the origin
due to the entire horizontal rod.
1.
v
E
rod 1
=
k
e
λ d
ℓ
ˆ
ı
2.
v
E
rod 1
=
k
e
λ ℓ
d
2
ˆ
ı
3.
v
E
rod 1
=
−
k
e
λ d
ℓ
ˆ
ı
4.
v
E
rod 1
=
k
e
λ ℓ
d
ˆ
ı
5.
v
E
rod 1
=
−
k
e
λ ℓ
d
ˆ
ı
6.
v
E
rod 1
=
k
e
λ d
ℓ
(
ℓ
+
d
)
ˆ
ı
7.
v
E
rod 1
=
−
k
e
λ d
ℓ
(
ℓ
+
d
)
ˆ
ı
8.
v
E
rod 1
=
−
k
e
λ ℓ
d
(
d
+
ℓ
)
ˆ
ı
correct
9.
v
E
rod 1
=
k
e
λ d
ℓ
2
ˆ
ı
10.
v
E
rod 1
=
k
e
λ ℓ
d
(
d
+
ℓ
)
ˆ
ı
Explanation:
±or the entire rod, we must add all the
segments,
i.e.
, integrate From
x
=
d
to
x
=
d
+
ℓ
. The electric feld due to the horizontal
rod is
v
E
rod 1
=
i
d
v
E
=
−
k
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 Spring '08
 Turner
 Charge, Work

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