A proton moves at 3.55 105 m/s in the horizontal direction. It enters a uniform

Question

A proton moves at 3.55 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 6.25 103 N/C. (Ignore any gravitational effects in your calculations.)
(a) Find the time interval required for the proton to travel 9.50 cm horizontally.
?t = 3 s

(b) Find its vertical displacement during the time interval in which it travels 9.50 cm horizontally.
h = 4 m

(c) Find the vertical component of its velocity after it has traveled 9.50 cm horizontally.
vy = 5 m/s

(d) Find the horizontal component of its velocity after it has traveled 9.50 cm horizontally.
vx = 6 m/s

Explanation / Answer

a) Since the field is vertical, the only force is vertical and the horizontal velocity does not change. Therefore t =d/vh = 0.095/(3.55*10^5) s=0.0267 * 10^-5 sec b) the vertical displacement in time t is 0.5*a*t²; t you have from part a), and acceleration a is F/m = qe*E/mp qe = elemental charge, mp = proton mass =1.6*10^-19 *6.25 * 10^3/9.11 * 10^-31 =1.097 * 10^15 c) v = a*t, a and t as computed above. =>v=1.097 * 10^15 * 0.0267 * 10^-5 =2.92 * 10^9 d) see a) hence

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