Question

002 (part 1 of 3) 10.0 points A proton has an initial velocity of 3.62 x 107 m/s in the horizontal direction. It enters a uniform electric field of 21500 N/C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0.056 m horizontally. The mass of the proton is 1.6726 x 10-27 kg and the fundamental charge is 1.602 x 10-19 C. Answer in units of ns. 003 (part 2 of 3) 10.0 points What is the vertical displacement of the pro- ton after the electric field acts on it for that time? Answer in units of mm. 004 (part 3 of 3) 10.0 points RE

Answer 1

E=21500N/C Som V=3.62x10mm -0.056m. D 7002 to o.osbus Time taken travell horizontally t = p whese, va initial horizontal speed = 3.62 x land . (ax=0) D=0.osom st=0.osom 3.62 X167m's t =1.55X10°sec or Italissus --- Answer

proton 18003 Net vertical force on Fnet =qE may=qE QE "ay where, q = change on proton = te = 1.602xcolc E=20soow/c, m=1.672 6 8 10 27 kg ay = vertical accele catim = ? ian - (16602 X10'9)(21500N/c) -E2.06X10M2 1.6726x 1027kg → Now, ventical displacement during timet d= vytt 2 ayt? Where vy= initial vestical speed=0 id=0++(2.06x10msy) (1.55x15?s)? = 2.47X1oom = 0.00247 X10?m a=0.00247 mm 1004 ventical speed at timet after vý = Vytayt 20+ (2.06X10*2m/s) (1.55x15?s) = 3.1938 10² m/s total speed after time t v= Jw+ vya = 112.62 xíómp)7 (3.193x10m/s) = 3620x10mls [V = 3620km/