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15_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

15_Chapter 15 HomeworkCH15 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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