we find F= 1. 78 *10 - 1 N in the negative y-direction b ) F = ma a = f = 1. 78 x10- N - 1, 95 x10'4 m/s ? m 9:11 X 10-31 Kg acceleration . is in the same direction as the net force. "and a = 1, 95 x 10 " m/s ? in the negative y - direction5 BOOK PAG TITLE PROJECT DATE Continued from Page Explain hand write out the problem- solving strategy for this problem. An electron is near a positive i'm of charge the and a negative ion of charge - se. See Heure below. - ge a ) find the magnitude and direction 10 of the resuttant in the electron . 4.00 MM b ) Find the magnitude and direction of the electure's instantaneous acceleration. OK 4.50 um electron 15 Solution 60.0 81 = Charge in pritive ion - ge ga - charge in negative im = - 80 8. = charge election 20 11 rip = distance from g, to 9. =4:50 um = 4.50 x 10 in 28 - 8, to 9, = 3.00 um = 3, 00 * 10-6 m = angle r, makes with negative x - axis = 6010 0 m = mass cy electron = 9:11 *10 - 31 kg 25 a ) Fig = K /8, 1 18:1 = (8. 99 *10 N-M / (2) (9 * Libx10 1)(16x10 2 ) ( 4:50m x 10 - 6mu ) a F3 = 1,02 X 10 - 16 N 30 F . = K 2, 1 16, ) = (8. 9 9 x10 ' N. m3 /c 2) ( 8 X 1.6 x 10 ( ) ( 1.6 *10- 12) ( 3 + 0 0 * 10 - m ) ? +23 = 2.05 x10 - 16 N ] Fx = ligx + Easy = - F,3 + E (60 - -102 X10 + ( 2 05x10 " N) 35 as 60 0 SIGNATURE Fy = Hoyt =0 - Fagsin60 = 0- 1. 05X10 / Sin60in po to Page TOATE Fy = - 1. 78 x10-To N