P22_014.PDF
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Chapter 22 - 22.14
14. (a) We choose the coordinate axes as shown in the diagram below. For ease of presentation (of the
computations below) we assume
Q>
0and
q<
0 (although the final result does not depend on
this particular choice). The repulsive force between the diagonallyopposite
Q
’s is along our (tilted)
x
axis. The attractive force between each pair of
Q
and
q
is along the sides (
o
f length
a
). In our
itself is therefore of length 2
d
=
a
√
2.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
x
q
•
a
•
Q
d
a
d
Q
•
•
q
Since the angle between each attractive force and the
x
axis is 45
◦
(note: cos 45
◦
=1
/
√
2)
, then the
net force on
Q
is
F
x
=
1
4
πε
0
(
Q
)(
Q
)
(2
d
)
2
−
2
(
|
q
|
)(
Q
)
a
2
cos45
◦
Q
2
2
a
2
−
=
1
4
πε
0
2
|
q
|·
Q
1
√
2
a
2
which (upon requiring
F
x
=0)leadsto
|
q
|
=
Q/
2
√
2or
q
=
−
Q
2
√
2
.
(b) The net force on
q
, examined along the
y
axis is
F
y
=
1
4
πε
0
q
2
(2
d
)
2
−
2
(
|
q
|
)(
Q
)
a
2
sin45
◦
q
2
2
a
2
−
=
1
4
πε
0
2
|
q
|·
Q
1
√
2
a
2
2
Q
√
2 which is inconsistent with the result of part (a).
Thus, we are unable to construct an equilibrium configuration with this geometry, where the only
forces acting are given byEq. 22-1.
−
drawing, the distance between the
c
enter to the corner is
d
,where
d
=
a/
√
2, and the diagonal
y
which (if we demand
F
y
=0)leadsto
q
=
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Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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