Question

A proton moves at 3.80 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.20 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally. 118.42 Correct: Your answer is correct. ns (b) Find its vertical displacement during the time interval in which it travels 4.50 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 4.50 cm horizontally.

Answer 1

(a) The time interval required for the proton to travel 4.50 cm horizontally is

t = (4.50 cm) / (3.80 x 105 m/s) = 118.42 ns

(b) The vertical displacement of the proton during the time interval is

dY = (F × t2) / 2m = (9.20 x 103 N/C × (118.42 ns)2) / (2 × 1.67 x 10-27 kg) = 0.0033 mm

(c) The horizontal and vertical components of the velocity after the proton has traveled 4.50 cm horizontally are

Vx = 3.80 x 105 m/s

Vy = (F × t) / m = (9.20 x 103 N/C × 118.42 ns) / (1.67 x 10-27 kg) = 1.90 x 105 m/s