24CHAPTER 15 The total charge enclosed by the cylindrical gaussian surface is QAσ=, where is the charge density on the conducting surface. Hence, Gauss’s law gives 0AEA=∈or oE=∈15.47( )( )()29219212222-158.9910 N mC1.6010C57.5 N2.00 10meekqqkeFrr−×⋅×====×15.48(a) ()2221982108.99C1.6010C8.210N0.5310mee−−−××F×(b) ( )2ecea mvrFm, so ( )1086-310.5310m8.210N2.210 m s9.1110kgerFvm−−××⋅=××15.49The three contributions to the resultant electric field at the point of interest are shown in the sketch at the right. The magnitude of the resultant field is REEE=−++3Eeekqk3122223123eReqqqqEkrrrr32+=−++
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.