Question

(1 point) -180000j N/C exists between A uniform field E two plates of length L 3 cm, as shown in the figure above. A proton is fired at an angle of 40° to the x axis with an initial speed of 6500000 m/s a) Find its vertical coordinates as it emerges from the plate, assuming the proton will not hit any plate in the process. cm b) Find the angle at which it emerges. degrees to the x axis TE

Answer 1

Vi sinto in 7 Net force on profon during the mation is =qE downward direction. Here we are negleeting grawlfational T fonce ng, due to small mass of - Vouso proton) E:-180000; charge on proton &q= e = 1167 16-19 c . NIC = 3cm mass of proton = m = 1.67x10-27 kg no= 6500000 acceleration of proton a = E=qE=CE = 40° of othe 9 = 16x76-19X180000 00 m/s2 (my directions 1167x10-27 1.9: 17.74 X1012 m/s2 Now initial velou proton in - derection = lg = Voose Ux= 6500000x cosyo" = 4979288.88 m/s acceleration profon in x-direction = 9x=0 ? No force in x-direction I distance required to travel in x -direction to enit from plafe = Sx=1=3cm = 3x10 m let it is the time taken by proton to frane Se we can write Sx = Unt t t axt2 a 34162 = 49 79 288.88 x t + 1/ xoxp? on t = 3x102 - S=0.00000DDD 602-5 4979288.88 f 6.02% 10-9 s V = velocity of proton in a direction after travelling distance L le Si

2019-10-22 Nour Vu - Urtgat - 4x + bet - Une a. Vx = Ux = Voloso - 4979288.88 m/s 2 -component of nelocity in r-direction remains Same, as there is no fonce in x-direction. puitial nelouty of profon iny-direction uy - M, Sh g = 65 евроох n o m/s Uy = 4178119.46 mle acceleration of profon in yo direction ay = -9 - 17:2481012 m/s2 time t = 61028 16-15 distance fronelled by proton in time & Cie. when proton comes out from the plate) = sy we can write Sy = lyft tayt? ch Sy = 414811914646002710-9+1 (-17.248 10") (60027409) m Sy = 0.0248 m = 2.48cm by a velocity of profon in yo direction at time of Eie. when comes out from plafe) My = ay tay t = 4178119.46-17-248101? x 61024/09 y = 4074334.66 m/s - tano = vy Ve de fano: 4074334.66 x sy a tano - o= 4979288.88 0.81825 tant (0.81825) = 39.29°

vertical co-ordinate of proton when it emerges from plate = Sy z 2:48cm 6 of of emerges at angle o emerges at angle o=39.29" from x