An electron is projected with an initial speed v 0 = 1.00×10 6 m/s into the unif

Question

An electron is projected with an initial speed v0 = 1.00×106 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.

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?

Explanation / Answer

A) Initial vertical velocity of the electron is zero.

Initial horizontal velocity of the electron is 1.0 x 10 m/s

Time taken to cross the plates = 0.02 / (1.0x10) = 20ns

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(2*10) + 1/2 a (2*10)²

a = 0.005 x 2 / (10¹) = 2.5*10¹3 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 2.5*10¹3 / (1.6x10¹) = 142.1875V/m

C)

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) (142.1875) / (1.67x10²) = 13.62 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(2*10) + 1/2 (13.62x10) (2*10)² = 2.724 x 10 m

The displacement is in the downward direction.

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