Introductory Physics Volume Two

1.7 More Examples 27
out the net force on the leftmost charge:
F L = 1 (4q)(q)
4πɛ 0 a 2 − 1 (4q)(4q)
4πɛ 0 (2a) 2
= 1 (4q)(q)
4πɛ 0 a 2 − 1 (4q)(q)
4πɛ 0 a 2
= 0
So the net force on the leftmost charge is zero. Since the rightmost
charge is the mirror image of the leftmost charge, it also has no net
force acting on it. Thus, all three charges have a net force of zero acting
on them for the configuration given, and the system is in equilibrium.
Is the equilibrium stable?
By stability we mean the following. If we move one of the charges
a little bit, does the configuration return to the original equilibrium
configuration, or does the configuration fall apart? Let’s move one of
the charges a little bit and see what happens. Move the left most charge
a small distance ɛ to the left. If the configuration is stable, then the
net force on it must push it to the right. Check:
a+ε
a
The net force is now:
+4q -q
+4q
ε
F L ′ = 1 (4q)(q)
4πɛ 0 (a + ɛ) 2 − 1 (4q)(4q)
4πɛ 0 (2a + ɛ) 2
If this force points to the right, it will push the charge back from
where it came and the equilibrium can be stable. If the force is to the
left, the charge will be pushed away from the equilibrium position and
the equilibrium can not be stable. With a little algebraic muscle the
formula above for F L ′ can be rewritten:
(
)
F L ′ =
q2 1
πɛ 0 a 2 (1 + ɛ − 1
a )2 (1 + ɛ
2a )2
It is easy to see that this is a negative number since the denominator
in the subtracted term is smallest, no matter how small ɛ is, as long
as it is larger than zero. Thus, the force on the leftmost charge points
to the left and the charge is pushed away from the original equilibrium
position. The equilibrium is unstable.
⊙ Do This Now 1.5