An electron is projected Figure E21.33 with an initial speed Vo 1.60 X 2.00 cm 1

Question

An electron is projected Figure E21.33 with an initial speed Vo 1.60 X 2.00 cm 106 m/s into the uniform field between the parallel plates in Fig vo E21.33. Assume that the field between 1.00 cm the plates is uniform and directed ver tically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. (a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. (b) Suppose that in Fig. E21.33 the electron is replaced by a proton with the same initial speed vo. Would the proton hit one of the plates? If the proton would not hit one of the plates, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (c) Compare the paths traveled by the electron and the proton and explain the differences. (d) Discuss whether it is reasonable to ignore the effects of gravity for each particle.

Explanation / Answer

as the charge of an electron is negative, force on it due to the electric field will be in opposite direction to the direction of the electric field.

let the magnitude of electric field strength=E

then force on the electron=q*E

acceleration=q*E/m

where q=charge on electron=1.6*10^(-19) C
m=mass of electron=9.1*10^(-31) kg.

as the horizontal speed is 1.6*10^6 m/s and the electric force does not affect this speed(because the electric field direction, hence the electric force direction is along vertical direction) .

time taken to travel 2 cm=distance/speed=12.5 ns

during this time, vertically the elctron has travelled 0.5 cm

so 0.5*acceleration*time^2=0.5 cm

putting all the values, we get E=364 V/m

part 2:

if we replace the electron with proton, due to higher mass of proton, the acceleration will be lesser as compared to electron.

hence vertical distance travelled will be less as compared to electron.

hence the proton wont hit the plate.

c) as proton is of positive charge, the force on it will be along the direction of electric field

so it will travel in the oppoiste direction of electron (vertically, i.e. it will go down as opposed to electron's upward movement)

d)as gravitational force is given as mass*9.8, here force will be of the order 10^(-30) for electron and 10^(-26) for proton. so they are negligible as compared to electric forces.

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