An electron is projected with an initial speed v0 = 1.40×106 m/s into the unifor

Question

An electron is projected with an initial speed v0 = 1.40×106 m/s into the uniform field between the parallel plates in the figure (Figure 1) . Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. If the proton would not hit one of the plates, what would be the magnitude of its vertical displacement as it exits the region between the plates?

For the magnitude of electric field I have 279 N/C
I keep getting 2.087*10^(-6) m for my vertical displacement on mastering physics and it says it's wrong. I tried every variation by now. Am I doing something incorrectly?

Explanation / Answer

Question a

Initial vertical velocity of the electron is zero.

Initial horizontal velocity of the electron is 1.4 x 10 m/s

Time taken to cross the plates = 0.02 / (1.4x10) = 14.28ns

Given that the electron enters midway between the plates and it just misses the upper plate as it emerges. Hence the vertical displacement of the electron = 0.5cm = 0.005m

The acceleration of the electron in the vertical direction is calculated as follows.

s = ut + 1/2 at²

0.005 = 0(14.289) + 1/2 a (14.289)²

a = 4.9013 m/s²

Thus, the force acting on the electron due to the electric field is F = eE = ma

E = ma/e = 9.1 x 10³¹ x 4.90*1013 / (1.6x10¹) = 278.91V/m
________________________________

Question b

When a proton enters the field, the same magnitude of force acts on it but in the vertical direction.

Acceleration of the proton in the vertically direction

a = eE/m = (1.6x10¹) (279) / (1.67x10²) = 26.73 x 10 m/s²

The proton takes same time as the electron to cross the plates (assuming the proton has the same initial horizontal velocity as the electron)

Vertical displacement s = ut + 1/2 at² = 0(14.28*10-9) + 1/2 (26.73x10) (14.28*10-9)² = 2.7253 x 10 m

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