(17%) Problem 2: The figure shows an electron passing between two charged metal
ID: 1773168 • Letter: #
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(17%) Problem 2: The figure shows an electron passing between two charged metal plates that create a 98 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.) The initial speed of the electron is 3.1 x 106 m/s, and the horizontal distance it travels in the uniform field is 3.2 cm. Otheexpertta.com 33% Part (a) What is its vertical deflection in m? 33% Part (b) What is the vertical component of its final velocity in m/s? 33% Part (c) At what angle does it exit in degrees? Neglect any edge effects Grade Summary 0% 100% Potential sin0 cotan) tan() coso asin0 atan acotan) sinh0 Attempts remaining: (5% per attempt) detailed view acoso cosh0 tanho cotanh0 ODegrees Radians I give up! Hints: deduction per hint. Hints remaining Feedback: deduction per feedback. Submission History Hints Feedback Totals Totals 0% 10%Explanation / Answer
a) The time the electron travels in the electric field is:
t = D/Vx = .032m/3.1 x 10^6 sec = 1.03e-8 sec
E = 98 N/C
The vertical deflection is:
y = -(.5 * e * E * t^2)/m = .00091 m
b) The vertical component of the velocity is:
Vy = - eE/m *t = -1.78 x 10^5 m/sec
c) Angle = arctan(vy/vx) = arctan(1.78x10^5/3.1x10^6) = 3.28 degrees
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