pasha (sep635) – HW 33 – Antoniewicz – (56445)
1
This
printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 4.0 points
A moving electron passes near the nucleus of
a gold atom, which contains 79 protons and
118 neutrons.
At a particular moment, the
electron is a distance of 4
.
7
×
10

9
m from
the gold nucleus.
What is the magnitude of
the force exerted by the gold nucleus on the
electron?
Correct answer: 8
.
22835
×
10

10
N.
Explanation:
The magnitude of the electric force is given
by
F
=
1
4
π
0
q
1
q
2
r
2
.
Plugging in the appropriate values, we obtain
F
=
1
4
π
0
(79
e
)(
e
)
(4
.
7
×
10

9
m)
2
=
8
.
22835
×
10

10
N
.
where
e
is the fundmental unit of charge
1
.
6
×
10

19
C.
002 (part 2 of 2) 4.0 points
What is the magnitude of the force exerted by
the electron on the gold nucleus?
Correct answer: 8
.
22835
×
10

10
N.
Explanation:
By Newton’s third law, interacting objects
must exert equal and opposite forces on each
other. Therefore the force on the gold nucleus
due to the electron is the same as the force on
the electron due to the gold nucleus.
003
8.0 points
The figure below shows two positively charged
objects and one negatively charged object.

+
+
What
is
the
direction
of
the
net
elec
tric force on the negatively charged object?
Choose your answer from the diagram labeled
IIX.
I
II
III
IV
V
VI
VII
VIII
IX — zero net force
1.
I
2.
V
3.
VIII
4.
IX
5.
II
6.
IV
7.
VII
correct
8.
III
9.
VI
Explanation:
We can take advantage of symmetries in
this problem.
With respect to the vertical
axis, the two positive charges on the left are
equidistant from the negative charge on the
right, and since they are on opposite sides
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pasha (sep635) – HW 33 – Antoniewicz – (56445)
2
of the negative charge, their vertical (or
y
)
components will cancel out. That leaves their
horizontal (or
x
) components.
So the force
will either point to the right or to the left
(along the +
x
or

x
axis, respectively). Given
that the charges on the left are positive, the
force must be attractive, so the net electric
force points toward the left.
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 Fall '08
 Turner
 Force, Correct Answer, Electric charge, pasha

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