Question

8. (-/3 Points] DETAILS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER The figure below shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) YA + + + + + x e VO The initial speed of the electron is 8.00 x 106 m/s, and the horizontal distance it travels in the uniform field is 2.00 cm. (a) What is its vertical deflection in meters? m (b) What is the vertical component of its final velocity in meters per second? m/s (c) At what angle a does it exit? Neglect any edge effects.

Answer 1

Solution in the uploaded image

Solution + Vi Fe=CF E=100N/c Vo = 8x10 m/s d = 2cm vo b ty E d Accelaration in anis is, y ay = er her acceleration in x anis is zero. an-o vo 2.5x10 sec Time taken to cover horizontal distance, to d t= 2x10-2 8x106 using equation of motion in y awes, y = Lloy at +1 ay 7² y = 0+ e (100) Ź (en 2)(2.5208) ly = 5:49x10 me -5 m + ayt Uy = me = ote(lo) 2.5x10-7 y = 43970.5 m/s Angle, &- by tent of a tent (43970.5 (0) 0.315°

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