A proton moves at 3.60 105 m/s in the horizontal direction. It enters a uniform

Question

A proton moves at 3.60 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.00 103 N/C. Ignore any gravitational effects.
(a) Find the time interval required for the proton to travel 4.00 cm horizontally.
? ns

(b) Find its vertical displacement during the time interval in which it travels 4.00 cm horizontally. (Indicate direction with the sign of your answer.)
? mm

(c) Find the horizontal and vertical components of its velocity after it has traveled 4.00 cm horizontally.
[v with arrow] =  ? [i] + ? [j] km/s

Explanation / Answer

a)
The horizontal motion is unaffected by the electric field eventhough it will change its direction.
Initial velocity = 3.60 x 105 m/s
Distance traveled = 0.04 m
Time taken = 0.04 / (3.60 x 105) = 0.0111 x 10-5 s.

b)
The electric force exerted by the field = qE
q is the charge of the proton
F = (1.6 × 10-19) x (8.00 x 103) = 12.8 x 10-16 N
Acceleration, a = F/m
m is the mass of proton, m = 1.675 × 10-27 kg
a = (12.8 x 10-16) / (1.675 × 10-27) = 7.64 x 1011 m/s2.

Initial vertical velocity = 0
displacement = 1/2 at2 = 0.5 x (7.64 x 1011) x (0.0111 x 10-5)2.
= 4.7 mm

c)
Vertical velocity = at
= (7.64 x 1011) x (0.0111 x 10-5) = 0.85 x 105 m/s
v = 3.60 x 105 m/s [i] + 0.85 x 105 m/s [j] = 360 [i] + 85 [j] km/s

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