24_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

24_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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24 CHAPTER 15 The total charge enclosed by the cylindrical gaussian surface is QA σ = , where is the charge density on the conducting surface. Hence, Gauss’s law gives 0 A EA = or o E = 15.47 ( )( ) () 2 92 1 9 2 12 2 22 -15 8.99 10 N m C 1.60 10 C 57.5 N 2.00 10 m e e kqq ke F rr ×⋅ × == = = × 15.48 (a) ( ) 2 2 2 1 9 8 2 10 8.99 C 1.60 10 C 8.2 10 N 0.53 10 m e e × × F × (b) ( ) 2 ec e a mvr Fm , so ( ) 10 8 6 -31 0.53 10 m 8.2 10 N 2.2 10 m s 9.11 10 kg e rF v m −− ×× = × × 15.49 The 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 R EE E =− + + 3 E ee kq k 3 1 2 2 2 2 3 1 2 3 e Re q q q q Ek r r r r 3 2 + = + +
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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.

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