CH 18 HW Begin Date: 1/22/2017 12:00:00 AM Due Date: 3/21/2017 11:59:00 PM End D

Question

CH 18 HW Begin Date: 1/22/2017 12:00:00 AM Due Date: 3/21/2017 11:59:00 PM End Date: 5/12/2017 12:00:00 AM Cl9%) Problem 13: The figure shows an electron passing between two charged metal plates that create a 110 NIC 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.) The initial speed of the electron is 2.5 x 10 m/s, and the horizontal distance it travels in the uniform field is 4.8 cm. 33% Part (a) What is its vertical deflection in m? A 33% Part (b What is the vertical component of its final velocity in m/s? A 33% Part co At what angle does it exit in degrees? Neglect any edge effects Otheexpertta.com Grade Summary Deductions 100% Potential

Explanation / Answer

let x-axis be the horizontal and y -axius be the vertical distance

since there is field component along horizontal direction

then ax = 0 m/sec

and ay = Fy/m = q*E/m = (1.6*10^-19*110)/(9.11*10^-31)

ay = 1.93*10^13 m/s^2

the time taken along x-axis to go completely out of the plates is t

then x = Vox*t

0.048 = 2.5*10^6*t

t = 1.92*10^-8 sec

verical distance travelled is y = (Voy*t)+(0.5*ay*t^2)

initial vertical component of velocity is Voy = 0 m/sec

y = (0*t)+(0.5*1.93*10^13*(1.92*10^-8)^2) = 0.003557 m

using Vy = Voy + (ay*t) = 0 +(1.93*10^13*1.92*10^-8)

Vy = 3.705*10^5 m/sec

tan(theta) = y/x = 0.003557 / 0.048 = 0.0741

theta = tan^(-1)(0.0741) = 4.23 degrees

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