An electron is accelerated inside a parallel plate capacitor. The electron leave

Question

An electron is accelerated inside a parallel plate capacitor. The electron leaves the negative plate with a negligible initial velocity and then after the acceleration it hits the positive plate with a final velocity beta. The distance between the plates is 17.6 cm, and the voltage difference is 140 kV. Determine the final velocity beta of the electron using classical mechanics. (The rest mass of the electron is 9.11 times 10^-31 kg, the rest energy of the electron is 511 keV.) 2 217e8 What is the final velocity beta of the electron if you use relativistic mechanics?

Explanation / Answer

d = 17.6 cm = 0.176 m ; V = 140 kV = 140 x 10^3 V ;

a)From conservation of energy:

1/2 m b^2 = q V

b = sqrt (2 q V/m)

b = sqrt (2 x 1.6 x 10^-19 x 140 x 10^3/9.1 x 10^-31) = 2.22 x 10^8 m/s

Hence, b = 2.22 x 10^8 m/s, using classical mechanics.

b)E0 = 511 keV = 511 x 10^3 x 1.6 x 10^-19 J = 8.18 x 10^-14 J

total relativistic energy is:

Et = E0 + K = E0 + mc^2 - m0c^2

k = m0c^2/sqrt (1 - v^2/c^2) - m0c^2

k = m0c^2 [1/sqrt(1 - v^2/c^2) - 1]

from work energy theroem,

W = Kef - KEi

W = m0c^2 [1/sqrt(1 - vf^2/c^2) - 1] -m0c^2[1/sqrt(1 - vi^2/c^2 - 1)]

W = m0c^2 [1/sqrt(1 - b^2/c^2) - 1] - 0

q V = m0c^2 [1/sqrt(1 -b^2/c^2) - 1]

[1/sqrt(1 -b^2/c^2) - 1] = q V/ m0c^2

[1/sqrt(1 -b^2/c^2) - 1] = 1.6 x 10^-19 x 140 x 10^3/8.18 x 10^-14 = 0.274

1/sqrt(1 - b^2/c^2) = 1 + 0.274 = 1.274

sqrt (1 - b^2/c^2) = 0.785

1 - b^2/c^2 = 0.886

b^2/c^2 = 1 - 0.886 = 0.114

b = 0.338 c

b = 0.338 x 3 x 10^8 = 1.014 x 10^8 m/s

Hence, b = 1.014 x 10^8 m/s ; using relativistic mechanics.

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