An electron is projected with an initial speed v 0 = 1.20×10 6 m/s into the unif

Question

An electron is projected with an initial speed v0 = 1.20×106 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.

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 B

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 1.20 x 10 m/s

Time taken to cross the plates = 0.02 / (1.20x10) = 16.66ns

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(16.66 *109) + 1/2 a (16.66*109)²

a = 0.005 x 2 / (16.66*109) 2 = 3.60*10^-23 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 3.60*10^-23 m/s² / (1.6x10¹) = 20.49 *10^-35 N/C

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) (20.49 *10^-35 N/C) / (1.67x10²) = 19.63 x 10 -27 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(16.66*10 -9) + 1/2 (19.63x10-27) (16.66*109)² = 2.7245 x 10-42 m

The displacement is in the downward direction.

_____

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