56 Electric PotentialP25.14Arbitrarily take V=0 at point P. Then (from Equation 25.8) the potential at the original position ofthe charge is −⋅=−EsELcosθ. At the final point a, VEL=−.Suppose the table is frictionless:KUif+=+af012212 2 00 103001 50160 000100030026−=−=−=×−°=−qELmvqELvqELmcoscos..cos...ejbgafC NCmkgmsSection 25.3Electric Potential and Potential Energy Due to Point ChargesP25.15(a)The potential at 1.00 cm is Vkqre1919278 99 101 60 10100 10144 10==×⋅××=×−−−..Nm CCmV22ej.(b)The potential at 2.00 cm is qre29278 99 101 60 10200 100719 10××−−−..CmVej.Thus, the difference in potential between the two points is ∆VV V=−=−×−8719 10. V.(c)The approach is the same as above except the charge is −×−160 1019C. This changes thesign of each answer, with its magnitude remaining the same.
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