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234
Chapter 23
(iii)
50.0 m
b
r
=
b
r
is now ten times larger than in (ii) so
U
b
is ten times smaller:
0.00992 J/10
0.000992 J.
b
U
=+
4
2(
)
2( 0.198 J
0.000992 J)
37.5 m/s.
2.80 10 kg
ab
b
UU
v
m
−
−+
−
==
=
×
EVALUATE:
The force between the two charges is repulsive and provides an acceleration to
q.
This causes the
speed of
q
to increase as it moves away from
Q.
23.8.
IDENTIFY:
Call the three charges 1, 2 and 3.
12
13
23
UU U U
+
SET UP:
12
23
13
UUU
because the charges are equal and each pair of charges has the same separation, 0.500 m.
EXECUTE:
26
2
3
3 (1.2 10 C)
0.078 J.
0.500 m
0.500 m
kq
k
U
−
×
=
EVALUATE:
When the three charges are brought in from infinity to the corners of the triangle, the repulsive
electrical forces between each pair of charges do negative work and electrical potential energy is stored.
23.9.
IDENTIFY:
12
13
23
12
13
23
qq
Uk
rr
r
⎛⎞
+
⎜⎟
⎝⎠
SET UP:
In part (a),
12
0.200 m
r
=
,
23
0.100 m
r
=
and
13
0.100 m.
r
=
In part (b) let particle 3 have coordinate
x
, so
12
0.200 m
r
=
,
13
rx
=
and
23
0.200
.
=−
EXECUTE: (a)
7
(4.00nC)( 3.00nC) (4.00nC)(2.00nC) ( 3.00nC)(2.00nC)
3.60 10 J
(0.200 m)
(0.100 m)
(0.100 m)
−
−−
+
=
×
(b)
If
0
U
=
, then
12
12
0.
k
r
x
+
−
Solving for
x
we find:
2
86
0
60
60
26
1.6
0
0.074 m, 0.360 m.
0.2
xx
x
+
−
⇒
−
+
= ⇒ =
−
Therefore,
0.074 m
x
=
since it is the only
value between the two charges.
EVALUATE:
13
U
is positive and both
23
U
and
12
U
are negative. If
0
U
=
, then
13
23
12
.
For
0.074 m
x
=
,
7
13
9.7 10 J
U
−
×
,
7
23
4.3 10 J
U
−
×
and
7
12
5.4 10 J.
U
−
×
It is true that
0
U
=
at this
x
.
23.10.
IDENTIFY:
The work done on the alpha particle is equal to the difference in its potential energy when it is moved
from the midpoint of the square to the midpoint of one of the sides.
SET UP:
We apply the formula
.
a
b
WU
U
→
In this case,
a
is the center of the square and
b
is the midpoint of
one of the sides. Therefore
center
side
center
side
.
U
→
There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a single
alphaelectron pair. At the center of the square, the alpha particle is a distance
r
1
=
50 nm from each electron. At
the midpoint of the side, the alpha is a distance
r
2
= 5.00 nm from the two nearest electrons and a distance
r
2
=
125 nm from the two most distant electrons. Using the formula for the potential energy (relative to infinity) of
two point charges,
00
(1/ 4
)(
/ ),
Uq
q
r
π
=
P
the total work is
center
side
center
side
U
→
=
01
02
03
11
1
42
2
44
4
ee
e
r
αα
α
ππ
PP
P
Substituting
q
e
= e
and
q
= 2
e
and simplifying gives
2
center
side
2 3
12 11
4
4
We
r
→
⎡
⎤
−
+
⎢
⎥
⎢
⎥
⎣
⎦
P
EXECUTE:
Substituting the numerical values into the equation for the work gives
()
2
19
21
211
4 1.60 10
C
6.08 10
J
5.00 nm
50 m
125 nm
W
−
−
⎡⎤
×
−
+
=
×
⎢⎥
⎣⎦
???
EVALUATE:
Since the work is positive, the system has more potential energy with the alpha particle at the center
of the square than it does with it at the midpoint of a side.
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235
23.11.
IDENTIFY:
Apply Eq.(23.2). The net work to bring the charges in from infinity is equal to the change in potential
energy. The total potential energy is the sum of the potential energies of each pair of charges, calculated from
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Charge, Acceleration, Force

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