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.)

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

Explanation / Answer

a)The work done by the electron to move between the plates is W = V * q where V = 113 kV = 113 * 10^3 V and q = 1.6 * 10^-19 C The work done is equal to the kinetic energy of the electron,therefore,we get W = K = (1/2)mß^2 or ß^2 = (2W/m) = (2V * q/m) or ß = (2V * q/m)^1/2 m = 9.1 * 10^-31 kg b)The rest energy of the electron is E = 511 keV or m * v^2 = E or v^2 = (E/m) or v = (E/m)^1/2 where E = 511 keV = 511 * 10^3 eV = 511 * 10^3 * 1.6 * 10^-19 J The final velocity ß of the electron if you use relativistic mechanics is ß = (v/(1 - v^2/c^2)^1/2) where c = 3 * 10^8 m/s

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