The speed of the electron as it emerges from the field is; 388.587 m/s
<h3>What is the speed of the electron?</h3>
Initial speed; v₀x = 1.1 * 10⁶ m/s
Acceleration in horizontal direction = 0 m/s²
distance; s_x = 2 cm = 0.02 m
Thus, formula to find time here is;
t = s_x/v₀x
t = 0.02/(1.1 * 10⁶)
t = 1.82 * 10⁻⁶ s
Now for the vertical distance; v,y_o = 0 m/s
Thus, the equation of motion becomes;
s_y = ¹/₂at²
0.005 = ¹/₂a(1.82 * 10⁻⁶)²
Solving for a gives;
a = 3.02 * 10¹³ m/s²
Thus the speed of the electron as it emerges from the field is;
v² = u² + 2as
v = √(0² + 2(3.02 * 10¹³ * 0.005))
v = 388.587 m/s
Read more about Electron speed at; brainly.com/question/15094100
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Complete Question is;
An electron is projected with an initial speed v0 = 1.1 * 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C
find the speed of the electron as it emerges from the field?.
