Assume that the field between the plates is uniform and directed vertically down

Question

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.
Part A If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. E = N/C 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.
Part A If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. E = N/C
Part C 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? |y| = m Part C 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? |y| = m

Explanation / Answer

Initial vertical velocity of the electron is zero.

Initial horizontal velocity of the electron is 2.0 x 10 m/s (Is that what you meant in the question or is it 2x106 = 212m/s?)

Time taken to cross the plates = 0.02 / (2.0x10) = 10ns

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(10) + 1/2 a (10)²

a = 0.005 x 2 / (10¹) = 10¹ 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 10¹ / (1.6x10¹) = 568.75V/m
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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 downward direction

a = eE/m = (1.6x10¹) (568.75) / (1.67x10²) = 54.49 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(10) + 1/2 (54.49x10) (10)² = 2.7245 x 10 m

The displacement is in the downward direction.
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Question c

Direction is given as Tan¹(Vertical displacement / horizontal displacement)

You get it as -0.0078° with respect to horizontal

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