An electron is projected with an initial speed v 0 = 1.50×10 6 m/s into the unif

Question

An electron is projected with an initial speed v0 = 1.50×106 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 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?

|y| = m

Explanation / Answer

time taken to emerge from Field.

Parallel to plates,

t = (0.02 m ) / (1.50 x 10^6 m/ s)

t = 1.333 x 10^-8 sec

In this time, displacement of electron perpendicular to the plates,

d = u t + a t^2 /2

(0.01 / 2) = 0 + a (1.333 x 10^-8)^2 / 2

a = 5.625 x 10^13 m/s^2

(A) a = q E / m

5.625 x 10^13 = (1.602 x 10^-19) ( E) / (9.109 x 10^-31)

E = 319.84 N/C ......Ans

(C) a = (1.602 x 10^-19 ) ( 319.84) / (1.673 x 10^-27)

a = 3.0626 x 10^10 m/s^2

d = 0 + ( 3.0626 x 10^10) (1.333 x 10^-8)^2 / 2

d = 2.722 x 10^-6 m

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