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Chapter 22 - 22.9
9. (a) If the system of three charges is to be in equilibrium, the force on each charge must be zero. Let
the third charge be
q
0
. It must lie between the other two or else the forces acting on it due to the
other charges would be in the same direction and
q
0
could not be in equilibrium. Suppose
q
0
is a
distance
x
from
q
, as shown on the diagram below. The force acting on
q
0
is then given by
F
0
=
1
4
πε
0
qq
0
x
2
−
4
qq
0
(
L
−
x
)
2
where the positive direction is rightward. We require
F
0
= 0 and solve for
x
. Canceling common
factors yields 1
/x
2
=4
/
(
L
−
x
)
2
and taking the square root yields 1
/x
=2
/
(
L
−
x
). The solution
is
x
=
L/
3.
←−
x
−→←−−
L
−
x
−−→
•
•
•
q
q
0
4
q
The force on
q
is
qq
0
.
F
q
=
−
1
4
πε
0
x
2
+
4
q
2
L
2
The signs are chosen so that a negative force value would cause
q
to move leftward. We require
F
q
= 0 and solve for
q
0
:
4
9
q
where
x
=
L/
3 is used. We now examine the force on 4
q
:
4
qx
2
L
2
q
0
=
−
=
−
F
4
q
=
1
4
πε
0
4
q
2
L
2
+
4
qq
0
(
L
=
1
4
πε
0
4
q
2
L
2
+
4(
4
/
9)
q
2
(4
/
9)
L
2
−
x
)
2
4
q
2
L
2
−
L
2
=
1
4
πε
0
4
q
2
(4
/
9)
q
and
x
=
L/
3, all three charges are in equilibrium.
(b) If
q
0
moves toward
q
the force of attraction exerted by
q
is greater in magnitude than the force of
attraction exerted by 4
q
.Thiscauses
q
0
to continue to move toward
q
and away from its initial
position. The equilibrium is unstable.
−
−
which we see is zero. Thus, with
q
0
=
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