benavides (jjb2356) – homework 03 – Turner – (59130)
1
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001
10.0 points
The
diagram
shows
an
isolated,
positive
charge
Q
, where point
B
is twice as far away
from
Q
as point
A
.
+
Q
A
B
0
10 cm
20 cm
The ratio of the electric field strength at
point
A
to the electric field strength at point
B
is
1.
E
A
E
B
=
2
1
.
2.
E
A
E
B
=
4
1
.
correct
3.
E
A
E
B
=
1
1
.
4.
E
A
E
B
=
1
2
.
5.
E
A
E
B
=
8
1
.
Explanation:
Let :
r
B
= 2
r
A
.
The electric field strength
E
∝
1
r
2
, so
E
A
E
B
=
1
r
2
A
1
r
2
B
=
r
2
B
r
2
A
=
(2
r
)
2
r
2
=
4
.
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
π
2.
Δ
q
=
Q
Δ
θ
2
π
3.
Δ
q
=
Q
2
π
4.
Δ
q
=
2
Q
π
5.
Δ
q
= 2
π Q
6.
Δ
q
=
Q
Δ
θ
π
correct
7.
None of these
8.
Δ
q
=
π Q
9.
Δ
q
=
2
Q
Δ
θ
π
10.
Δ
q
=
Q
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
.
003
(part 2 of 3) 10.0 points
The magnitude of the
x
component of the
electric field at the center, due to Δ
q
, is given
by
1.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
2.
Δ
E
x
=
k

Δ
q

cos
θ
r
3.
Δ
E
x
=
k

Δ
q

(sin
θ
)
r
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benavides (jjb2356) – homework 03 – Turner – (59130)
2
4.
Δ
E
x
=
k

Δ
q

sin
θ
r
2
5.
Δ
E
x
=
k

Δ
q

(sin
θ
)
r
2
6.
Δ
E
x
=
k

Δ
q

r
2
7.
Δ
E
x
=
k

Δ
q

r
2
8.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
2
9.
Δ
E
x
=
k

Δ
q

sin
θ
r
10.
Δ
E
x
=
k

Δ
q

cos
θ
r
2
correct
Explanation:
Negative charge attracts a positive test
charge. At
O
, Δ
E
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 Spring '08
 Turner
 Work, Electric charge, kq

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