Unformatted text preview: )(2 qQ/h 2 ), at a distance L/ 2 from the bearing. This torque is positive. The equation for rotational equilibrium is − 1 4 πε qQ h 2 L 2 − W µ x − L 2 ¶ + 1 4 πε 2 qQ h 2 L 2 = 0 . The solution for x is x = L 2 µ 1 + 1 4 πε qQ 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 W − 1 4 πε qQ h 2 − 1 4 πε 2 qQ h 2 − N = 0 . We solve for h so that N = 0. The result is h = r 1 4 πε 3 qQ W ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge, Force

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