P22_017.PDF

17. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net torque about any

point is also zero. We write an expression for the net torque about the bearing, equate it to zero,

and solve for x . The charge Q on the left exerts an upward force of magnitude (1 / 4 πε 0 )( qQ/h 2 ),

at a distance L/ 2 from the bearing. We take the torque to be negative. The attached weight exerts

a downward force of magnitude W , at a distance x

−

4 πε 0

h 2

2 − W x −

4 πε 0

2 qQ

h 2

2 =0 .

The solution for x is

x = L

4 πε 0

h 2 W

(b) If N is the magnitude of the upward force exerted by the bearing, then Newton’s second law (with

zero acceleration) gives

4 πε 0

h 2 −

4 πε 0

2 qQ

h 2

−

N =0 .

We solve for h so that N = 0. The result is

h = 1

4 πε 0

3 qQ

L/ 2 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

−