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/2
x
L
−
from the bearing. This torque is also negative. The charge
Q
on the right exerts
an upward force of magnitude (1/4
πε
0
) (2
qQ
/
h
2
), at a distance
L
/2 from the bearing. This
torque is positive. The equation for rotational equilibrium is
22
00
11
2
0.
42
2
4
2
qQ L
L
qQ L
Wx
hh
εε
−
⎛⎞
−
−+
=
⎜⎟
π
⎝⎠
π
The solution for
x
is
x
Lq
Q
hW
=+
F
H
G
I
K
J
2
1
1
4
0
2
π
ε
.
(b) If
F
N
is the magnitude of the upward force exerted by the bearing, then Newton’s
second law (with zero acceleration) gives
2
44
N
qQ
qQ
WF
πε
−−
−
=
We solve for
h
so that
F
N
= 0. The result is
h
qQ
W
=
1
4
3
0
π
.
44. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net torque
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 Spring '08
 Any
 Physics, Charge, Force

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