002201.2 J44fqqqqKUUxyxyπε′′+=⇒+=++. This yields y= −8.5 m. 59. We apply conservation of energy for particle 3 (with q'= −15 ×10− 6C): K0+ U0= Kf+ Ufwhere (letting x= ±3 m and q1= q2= 50 ×10−6C = q) 1200 02444qqqqUxy′′′=+=+++. (a) We solve for Kf(with y0= 4 m): 002111.2 J3.0 J4||ffqqKK
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This note was uploaded on 12/13/2010 for the course PHYS 144 taught by Professor Ig during the Fall '10 term at NE Texas CC.