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 19.0 cm, and the voltage difference is 113 kV.

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

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

Answer 1

a) Workdone on the electron, W = F*d*cos(0)

= q*E*d

= q*V

= 1.6*10^-19*113*10^3

= 1.808*10^-14 J

we know, Workdone = gain in kinetic energy

W = (1/2)*m*v^2

==> v = sqrt(2*W/m)

= sqrt(2*1.808*10^-14/(9.1*10^-31))

= 1.99*10^8 m/s <<<<<<<<---------------------------Answer

b)
relativistic kinetic energy, KE = mo*c^2*(1/sqrt(1 - (v/c)^2) - 1)

W = mo*c^2*(1/sqrt(1 - (v/c)^2) - 1)

1.808*10^-14 = 9.1*10^-31*(3*10^8)^2*(1/sqrt(1 - (v/(3*10^8))^2) - 1)

on solving the above equation we get,

==> v = 1.72*10^8 m/s <<<<<<<<---------------------------Answer