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

Question

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 2.6 times 10^6 m/s, and the horizontal distance it travels in the uniform field is 3.2 cm. What is its vertical deflection in m? What is the vertical component of its final velocity in m/s? At what angle does it exit in degrees? Neglect any edge effects.

Explanation / Answer

Calculate Vertical Deflection -

So Here Given its velocity of 2.6*10^6 m/s and the distance of 3.2 cm (0.32 m)

So we can quickly determine "t," the time of travel

t = 0.32 / 2.6*10^6 = 1.230* 10-7 sec

Now Newton's second law = F= ma

a = F/m = E*e/m = (98 N/C) * (1.6*10^-19 C) / (9.11 * 10^-31 kg) =

a = 1.721 * 10^13 m/s/s

Now v (vertical)= a*t = 1.721 * 10^13 m/s/s * 1.230* 10-7 sec =

v (vertical) = 2.11 * 10^6 m/s

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