An ion with a positive charge equal to 3 times the charge of a proton and with a

Question

An ion with a positive charge equal to 3 times the charge of a proton and with a mass of 8.78×10^-26 kg is traveling horizontally at a speed of 3.12×10^5 m/s when it enters a region of space that has an electric field with a strength of 13.5 N/C pointing straight down. Neglect any effects of gravity in this problem.

a) How much time will it take the ion to travel 10.0 cm vertically downward? (Hint, consider the horizontal and vertical forces and motion separately.)
?seconds

b) How far will the ion have traveled horizontally during that time?
?meters

I don't know where to start please help!!

Explanation / Answer

You can get the force on the ion by F = qE. They give you E, and q is 3 times the proton charge. So

F = (3 * 1.602 * 10-19 C)(13.5 N/C) = 6.49 * 10-18 N

Now use F = ma, solved for a, to get a = F/m

a = (6.49 * 10-18 N)/(8.78 * 10-26 kg) = 7.39 * 107 m/s2

Now use basic kinematics, d = (1/2)at2, solved for t to get t = sqrt(2d/a)

t = sqrt((2 * 0.10 m)/(7.39 * 107 m/s2)) = 5.202 * 10-5 s which is the answer to part (a).

To get part (b), just use d = vt:

d = (3.12 * 105 m/s)(5.202 * 10-5 s) = 16.2 m

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