57. Our approach (based on Eq. 22-29) consists of several steps. The first is to find an approximatevalue of eby taking differences between all the given data. The smallest difference is between the fifth and sixth values: 18.08 ×10 –19C – 16.48 ×10 – 19C = 1.60 ×10–19C which we denote eapprox. The goal at this point is to assign integers nusing this approximate value of e: 191approx192approx193approx194approx195approx6.563 10Cdatum14.1048.204 10Cdatum25.13511.50 10Cdatum37.19713.13 10Cdatum48.21816.48 10Cdatum510.3010nenenenene−−−−−×=¡=×=¡=×=¡=×=¡=×=¡=196appeox197approx198approx199approx18.08 10Cdatum611.301119.71 10Cdatum712.321222.89 10Cdatum814.311426.13 10Cdatum916.3316nenenene−−−−×=¡=×=¡=×=¡=×=¡=Next, we construct a new data set (e1, e2, e3…) by dividing the given data by the
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.