Based on Example 22.8,

Recalculate a) the particle position (x, y) and b) Vyr at time t= 2.4 X 10'8 s using the necessary values he has calculated, c) use mass of electron (9. 11 x 10-31 kg) to calculate the kinetic energy of the particle based on the velocity magnitude of at the given time. An electron enters the region of a uniform electric field as shown in the figure, with v; = 3.00 x 106 m/s and E = 200 N/C. The horizontal length of the plates is . = 0.100 m. (A) Find the acceleration of the electron while it is in the electric field. The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates. (0, 0) X ( x, y) E V + + + + + + ++++++eE The electron undergoes a downward a, =- acceleration (opposite E), and its motion m is parabolic while it is between the plates. 1.60 x10-19 C )(200 N/C) - vi 9.1 1x 10 -31 kg a (0, 0) -3.51x1013 m/s2 X = (x, y) E + + + + + + + + + + + +(B) Assuming the electron enters the field at time t = 0, find the time at which it leaves the field. X f - X; The electron undergoes a downward X=X TVtat= acceleration (opposite E), and its motion x is parabolic while it is between the plates. 1-0 0.100 m vi t= (0, 0) 3.00x10' m/s X ( x, y) E = 3.33x10 8 s V + + + + + + ++ ++ + +The electron undergoes a downward (C) Assuming the vertical position of the acceleration (opposite E), and its motion is parabolic while it is between the plates. electron as it enters the field is y; = 0, what is its vertical position when it leaves the field? v.i (0, 0) x y f = Vit vitt-at ( x, V) E 2 V + + + + + + + + ++++ yf = 0+0+- -3.51x1013 m/s? )(3.33x10 8 s ) N =-0.0195 m=-1.95 cm