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 ß. The distance between the plates is 13.5 cm, and the voltage difference is 113 kV.

a) Determine the final velocity ß of the electron using classical mechanics. (The rest mass of the electron is 9.11×10-31 kg, the rest energy of the electron is 511 keV.)

I got this answer, 6.65 * 10^1 But cant figure out part b below.

b) What is the final velocity ß of the electron if you use relativistic mechanics?

Explanation / Answer

in relativity, kinetic energy = mc^2*(r - 1) = qV r - 1 = qV/(mc^2) = 113/511 = 0.221135 r = 1.221135 = 1/sqrt[1 - (v/c)^2]
so v/c = sqrt(1 - r^-2) = 0.574

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