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 14.2 cm, and the voltage difference is 146 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.)
7.559×10-1

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

Explanation / Answer

Let velocity =

V=voltage diff = 146 * 10^3 V

KE = kinetic energy

mo = Rest mass of Electron = 9.11×10-31 kg

Rest energy of Electron (mo* c^2) = 511Kev

1ev = 1.6 * 10^-19 J

Rest energy of Electron (mo* c^2) = 511 *10^3 * 1.6 * 10^-19 J = 8.176 * 10^-14 J

We know Kinetic Energy Gained by Electron inside a parallel Plate capacitor = V * Charge on electron = 146 * 10^3 * 1.6 * 10^19 = 2.336* 10^-14 J

Kinetc energy of electron is given by -

KE = (Y-1)moc^2

Where Y = 1/ sqrt(1 - (/c)2) -----------1

Solving for Y

2.384 * 10^-14  = (Y-1) * 8.176 * 10^-14

Y = 2.336 * 10^-14  / 8.176 * 10^-14 + 1

Y = 1.285

Substituting Value of Y in Equation 1

Y = 1/ sqrt(1 - (/c)2)

1 - (/c)2 = 1/y2

(/c)2 = 1- 1/y2

= sqrt( 1- 1/y2) c

= sqrt (1-1/1.285^2) * 3* 10^8 m/s

= 1.884 * 10^8 m/s or 0.628c , where c is velocity of light

Final velocity of the electron using relativistic mechanics = 1.884 * 10^8 m/s or 0.628c   

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.