/2xL−from the bearing. This torque is also negative. The charge Qon the right exerts an upward force of magnitude (1/4πε0) (2qQ/h2), at a distance L/2 from the bearing. This torque is positive. The equation for rotational equilibrium is 22001120.42242qQ LLqQ LWxhhεε−⎛⎞−−+=⎜⎟π⎝⎠πThe solution for xis xLqQhW=+FHGIKJ211402πε.(b) If FNis the magnitude of the upward force exerted by the bearing, then Newton’s second law (with zero acceleration) gives 244NqQqQWFπε−−−=We solve for hso that FN= 0. The result is hqQW=1430π.44. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net torque
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.