15_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

# 15_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

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Electric Forces and Electric Fields 15 15.19 We shall treat the concentrations as point charges. Then, the resultant field consists of two contributions, one due to each concentration. The contribution due to the positive charge at 3 000 m altitude is ( ) () 2 95 2 22 40.0 C Nm 8.99 10 3.60 10 N C downward C 1000 m e q Ek r +  ==×   The contribution due to the negative charge at 1 000 m altitude is ( ) 2 2 40.0 C 8.99 3.60 C e q r The resultant field is then 5 7.20 10 N C downward +− =+= × EE E GG G 15.20 (a) The magnitude of the force on the electron is F qE eE = = , and the acceleration is ( ) ( ) 19 13 2 31 1.60 10 C 300 N C 5.27 10 m s 9.11 10 kg ee Fe E a mm × === = × × (b) ( ) 13 2 8 5 0 0 5.27 10 m s 1.00 10 s 5.27 10 m s a t =+=+ × × = × vv 15.21 If the electric force counterbalances the weight of the ball, then
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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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