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 11.0 cm, and the voltage difference is 147 kV. Determine the final velocity 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.)

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

Explanation / Answer

here,

Assuming Notations :

v = velocity =
V = voltage diff
KE = kinetic energy
Y = gamma
m = mass
q = charge
1 ev = 1.6*10^-19 J

By Classical Method Final Velocity of electron will be
v = sqrt(-(2*q*V)/(m) )
v = sqrt(-2( -1.6*10^-19*147000 ) / (9.11*10^-31 ) )
v = 2.27*10^8 m/s

as 1 m/s = 3.335* 10^-9 c

V = 0.757 c =

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
By Relativistic Method Final Velocity of electron will be :
if speeds higher than > 1/10c (10% speed of light), we have to use relativity for accuracy.

as, KE = (Y-1)m*c^2

solving for gamma, Y
Y = (KE / m*c^2) + 1
Y = ( (511000 * 1.60*10^-19 ) / ( (9.11*10^-31)(3*10^8)^2 ) + 1
Y = 1.997

Therefore Velocity will be ,
v = sqrt( 1- (1/Y^2) )*c
v = sqrt( 1- (1/(1.997)^2) )*(3*10^8)
v = 2.59*10^8 m/s

as 1 m/s = 3.335* 10^-9 c

V = 0.864 c =

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