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 B. The distance between the plates is 17.4 cm, and the voltage difference is 148 kv. Determine the final velocity B of the electron using classical mechanics. (The rest mass of the electron is 9.11x10-31 kg, the rest energy of the electron is 511 keV.) Submit Answer Tries 0/12 What is the final velocity B of the electron if you use relativistic mechanics? Submit Answer Tries 0/12

Answer 1

148K V The work done by an electric field in moving charge from a region of potential Vi to the region of potential Vf is - W = q (V, - Vf) For electron moving from negative black to the positive plate W=(-e) cov - 148 kv) = 148 Kev W=(1.602x1019 ()+( 148x103 V) W = 2.378 X10th J By Wook energy theorem, the net work done on an object is equal to the kinetic energy of the object. Know - Kinitial = w The election started from hest So Kinétical = 0 mer² -0 = 2.371810-4 J U-2x2.377x10-'4J V me Using me = toll x10-31 kg we get change i

.761 U=2.282x10% mis B= = 2.282x108 m/s 3x108 m/s [ B = 0.761] In relativity the Kinetic energy is given by K= mc²(8-1) where 8 = 1 Initially the election was at rest, so its initial Kinetic energy is zero. By work energy theorem Kiral - Kinitial = W mec2(8-1)-0 = 148 kev 8 = 1 + 148 kev V - B 8=1t 148ev 511 Kev = 1.290 Using 8 11 73 ) B= V1 - 7 B=0.632