ch21-p044 - 44. (a) Since the rod is in equilibrium, the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
/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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online