Question

A proton moves at 5.20 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.40 103 N/C. Ignore any gravitational effects. (a) Find the time interval required for the proton to travel 4.50 cm horizontally.

Answer 1

86.5 ns

Explanation:

The speed in the original direction (horizontally) is unchanged by the vertical force the field exerts. The travel time is ...

time = distance/speed = (4.5×10^-2 m)/(5.20×10^5 m/s) = 8.65×10^-8 s

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An engineer would express this time using the SI prefix nano- for 10^-9. The time is 86.5 ns.