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An electron is projected with an initial speed v0 = 4.85×106 m/sinto the uniform field between the parallel plates in the figure(Figure 1) . 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.

Part A

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Explanation / Answer

So let us consider that the initial vertical velocity of the electron is zero.

Initial horizontal velocity of the electron is 4.85 x 106 m/s

Time taken to cross the plates = 0.02 / (4.85 x 106) = 4.12 ns

It is given in the question that electron enters midway between the plates and it just misses the upper plate as it emerges. So here midway means half of the distance.

So, 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 = 0t + 1/2 a *(4.12*10-8)2

a = 5.88*1014 m/s2

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

E = ma/e = 9.1 x 10-31 x 5.88*1014 / (1.6x10-19) =3344.6 V/m

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