An electron is accelerated in the uniform field E (E = 9.80×10 3 N/C) between tw

Question

An electron is accelerated in the uniform field E (E = 9.80×103 N/C) between two parallel plates shown in the figure. (The thickness of the plates themselves is negligible, despite how fat they might look in this picture.)

The electron starts from rest near the negative plate and passes through a tiny hole in the positive plate. In designing such a system, how far apart do you need to space the two plates such that the electron exits with a kinetic energy of 2.27×10-17 J?
(As we will see later in this course, this energy is just enough to generate light when the electron finally hits phosphors on a screen!)

Explanation / Answer

KE = .5 * m *v^2
v^2 = 2 KE / m
v^2 = 2 * 2.27x10^-17 J / 9.11x10^-31kg
v^2 = 4.98x10^13 m^2/s^2
v = 7.059 x10^6 m/s

F = E * q
F = 9.80x10^3 N/C * 1.60x10^-19C
F = 1.568x10^-15 N

a = F/m
a = 1.568x10^-15N / 9.11x10^-31kg
a = 1.7211x10^15 m/s^2

v^2 = 2 a x
x = v^2 / 2 a
x = 4.98x10^13 m^2/s^2 / 2 * 1.7211x10^15 m/s^2
x = 1.4467 cm ---------------------->>>>>>>>>>>>>>>>answer)

I HOPE I HAVE ANSWERED THE QUESTION CORRECTLY.

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