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SOLUTIONS, ASSIGNMENT 9 plus Problem 23.25
23.10.
I
DENTIFY
:
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.
S
ET
U
P
:
We apply the formula
.
a
b
a
b
W
U
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
.
W
U
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,
0
0
(1/ 4
)(
/ ),
U
qq r
π
=
P
the total work
is
center
side
center
side
W
U
U
→
=

=
0
1
0
2
0
3
1
1
1
4
2
2
4
4
4
e
e
e
q q
q q
q q
r
r
r
α

+
P
P
P
Substituting
q
e
= e
and
q
= 2
e
and simplifying gives
2
center
side
0
1
2
3
1
2
1
1
4
4
W
e
r
r
r
→
= 

+
P
E
XECUTE
:
Substituting the numerical values into the equation for the work gives
(
29
2
19
21
2
1
1
4 1.60 10
C
6.08 10
J
5.00 nm
50 m
125 nm
W


= 
×

+
=
×
???
E
VALUATE
:
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.
23.14.
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 Fall '10
 dfs
 Energy, Potential Energy, Work

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