Electric Forces and Electric Fields1515.19 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 ( )()29522240.0 CNm8.9910 3.6010 N C downwardC1000 meqEkr+⋅==×=×The contribution due to the negative charge at 1 000 m altitude is ( )2240.0 C8.993.60Ceqr−⋅The resultant field is then 57.2010 N C downward+−=+=×EE EGG G15.20(a) The magnitude of the force on the electron is FqEeE==, and the acceleration is ( )( )19132311.6010C300 N C5.2710m s9.1110kgeeFeEamm−−×====××(b) ( )13285005.2710m s1.0010s5.2710 m sat−=+=+××=×vv15.21If the electric force counterbalances the weight of the ball, then
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