Question

4. A proton moves at 4.50x10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60x10^3 N/C. Ignoring any gravitational effects, find

a) the time interval required for the protons to travel 5.00 cm horizontally

b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally

c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally

Answer 1

a)

time, t = distance travelled / speed

t = 0.05/ (4.5*10^5) = 1.11*10^-7 s

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b)

Vertical acceleration

ma = qE

a = qE/m = 1.6*10^-19* 9.6*10^3/ ( 1.67*10^-27)

a = 9.198*10^11 m/s^2

using 2nd equation of motion

y = 0.5 at^2

y = 0.5* 9.198*10^11* (1.11*10^-7)^2

y = 5.666*10^-3 m

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c)

Vertical velocity of proton

Vy = at = 9.198*10^11* 1.11*10^-7

Vy = 1.02*10^5 m/s

Horizontal velocity of proton

Vx = 4.5*10^5 m/s

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