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 . The distance between the plates is 15.3 cm, and the voltage difference is 115 kV. Determine the final velocity of the electron using classical mechanics. (The rest mass of the electron is 9.11×10-31 kq, the rest energy of the electron is 511 keV.) 2.010e8 _Tries 3/12e what is the final velocity of the electron if you use relativistic mechanics?

Explanation / Answer

v= velocity = , V=voltage diff, KE = kinetic energy, Y = gamma, m = mass, and q = charge

Classical Method:

v = sqrt(-2qV/m) use this equation and solve for v. It is derived from the fact that KE = Vq

so, v = sqrt[-2(-1.6x10^-19)(115000)/ (9.11x10^-31 )] = 2x10^8 m/s = (0.67c)  

Relativistic Method:

At speeds higher than > 1/10c (10% speed of light), we have to use relativity for accuracy.

KE = (Y-1)mc^2 use this eqn to find gamma (Y) and use v = sqrt[1-(1/Y^2)]c to find the velocity.

Convert eV to Joules using the fact that 1 ev = 1x10^-19 J

Y = [(115000ev)(1.60x10^-19 J/ev)]/[(9.11x10^-19)(3x10^8)^2] + 1 = 1.224

v = sqrt{1-[1/(1.224)]}c = 1.28x10^8 m/s = (0.428c)

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