The figure below shows an electron passing between two charged metal plates that

Question

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.) The initial speed of the electron is 1.00 times 10^6 m/s, and the horizontal distance it travels in the uniform field is 5.00 cm. (a) What is its vertical deflection in meters? 044 m (b) What is the vertical component of its final velocity in meters per second? 8.8e5 m/s (c) At what angle theta does it exit? Neglect any edge effects. 5.03 degree

Explanation / Answer

In vertical direction

Force = Me *a = e*100 = 1.6 *10-19 *100

a = 1.6 *10-19 *100 / [9.1*10-31 ]

a = 1.76 *1013 m/sec (in vertical direction )

now time taken to pass this region

t = 5*10-2 / 106  = 5*10-8 sec

now

1) apply in vertical direction

s = ut +1/2 at2  

u = 0

s = 1/2 *1.76 *1013 *(5*10-8 )2 = 2.2 cm = 0.022 m

2) v = u +at = 0 +1.76 *1013 *5*10-8 = 8.8 *105

3) tanx = Vvertical / Vhorizontal

tanx = 8.8*105 / 1*106 = 0.88

x = 41.35 degree

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