An electron is projected with an initial speed Vo = 5.10 x 10^6 m/s into the uni

Question

An electron is projected with an initial speed Vo = 5.10 x 10^6 m/s into the uniform field between the parallel plates in the figure. The direction of the field is vertically downward, and the field is zero except in the space between the two plates. The electron enters the field at a point midway between the plates. The electron just misses the upper plate as it emerges from the field.

a) Express the magnitude of the electric field in terms of the acceleration of the electron (equations only - no numbers or units!)

b) Which way does the acceleration point?

c) How long does it take for the electron to leave the capacitor?

d) Find the magnitude of the acceleration.

e) Use parts (a) and (d) to find the magnitude of the electric field.

Explanation / Answer

a)

F = qE = ma

E = ma/q

b)

Acceleration of electron would be in vertical direction

c)

Initial vertical velocity of the electron is zero.

Initial horizontal velocity of the electron is 5. x 10 m/s

Time taken to cross the plates = 0.02 / (5.1x10) = 3.9 ns

d)

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 a

s = ut + 1/2 at²

0.005 = 0(3.9*109) + 1/2 a (3.9*109)²

a = 0.005 x 2 / (15.21*10¹8) =6.57* 10¹ m/s²

e)

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

E = ma/e = 9.1 x 10³¹ x 6.57* 10¹ / (1.6x10¹) = 37.3*10^2 V/m

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