An Emoktron (mass = 1.6 times 10^-27 Kg, charge = -1.6 times 10^-19 C) is accele

Question

An Emoktron (mass = 1.6 times 10^-27 Kg, charge = -1.6 times 10^-19 C) is accelerated in the uniform E-field (5 times 10^4 N/C) between two thin parallel charged plates. The separation of the plates is 2-cm. The Emoktron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate. (a) Setup the equation to find the speed when it's leavening the hole. (b) What will the E-field value have to be to have final speed of 9 times 10^7 m/s, leaving everything constant? (c) Giving the velocity of Emoktron in (b) calculate the relativistic mass? (d) Giving the velocity of Emoktron in (b) calculate the kenotic energy (KE) and momentum (hint relativistic)? (e) If the Emoktron at rest has lifetime of 1-second, what will new lifetime be (hint relativistic time)? (And if you think I misspelled Emoktron = Electron. No this is an emo-electron)

Explanation / Answer

a) we will use of energy,

electrostatic PE = Final KE

qEd = 1/2 mv^2

( where q = charge, E = electric field, d = distance between both plates, m = mass of emoktron. v= velocity at the hole)

b) qEd = 1/2 mv^2

1.6 x 10^-19 ( E) ( 2x 10^-2) = 0.5 (1.6 x 10^-27 ) (9 x 10^7) ^2

E ( 3.2 x 10^ -21 ) = (7.2 x 10^ -13)

E = 2.25 x 10^ 8 N/C

c) Mr ( relativistic mass) = 1.6 x 10^27 / sqroot ( 1-81 x 10^14/ 9 x 10^ 16)

Mr= 1.6 x 10^-27/ sqroot ( 1- 0.09)= 1.6 x 10^27/ 0.953 = 1.6789 x 10^- 27kg

d) KE = 1/2 ( 1.6789 x 10^ -27kg ) (81 x 10^14) = 67.995 x 10^ -13 J

P = mrV=  1.6789 x 10^- 27kg x 9 x 10^7 = 15.11x 10^ -20 kg m/s

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