15_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

15_Ch 15 College Physics ProblemCH15 Electric Forces and Electric Fields

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online