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tang (nmt352) – homework 03 – Turner – (56705)
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 feld strength at
point
A
to the electric feld strength at point
B
is
1.
E
A
E
B
=
1
1
.
2.
E
A
E
B
=
2
1
.
3.
E
A
E
B
=
4
1
.
correct
4.
E
A
E
B
=
1
2
.
5.
E
A
E
B
=
8
1
.
Explanation:
Let :
r
B
= 2
r
A
.
The electric feld 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 fgure 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
2.
Δ
q
= 2
π Q
3.
None oF these
4.
Δ
q
=
2
Q
π
5.
Δ
q
=
π Q
6.
Δ
q
=
Q
Δ
θ
2
π
7.
Δ
q
=
Q
8.
Δ
q
=
2
Q
Δ
θ
π
9.
Δ
q
=
Q
2
π
10.
Δ
q
=
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
.
003
(part 2 oF 3) 10.0 points
The magnitude oF the
x
component oF the
electric feld 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

(sin
θ
)
r
3.
Δ
E
x
=
k

Δ
q

sin
θ
r
2
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View Full Documenttang (nmt352) – homework 03 – Turner – (56705)
2
4.
Δ
E
x
=
k

Δ
q

r
2
5.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
6.
Δ
E
x
=
k

Δ
q

(cos
θ
)
r
2
7.
Δ
E
x
=
k

Δ
q

(sin
θ
)
r
2
8.
Δ
E
x
=
k

Δ
q

r
2
9.
Δ
E
x
=
k

Δ
q

cos
θ
r
2
correct
10.
Δ
E
x
=
k

Δ
q

sin
θ
r
Explanation:
Negative charge attracts a positive test
charge. At
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