Question

A tiny hole is made in the center of the negatively and positively charged plates of a capacitor, allowing a beam of electrons to pass through and emerge from the far side. If 59.2 V are applied across the capacitor plates and the electrons enter through the hole in the negatively charged plate with a speed of 9.7x103 km/s (note: 9.7 thousands kilo-meters per second), what is the speed of the electrons as they emerge from the hole in the positive plate? You may answer in km/s (or in m/s).

Answer 1

An electron's mass is 9.11 x 10^-31 kg, so at a velocity of 9.70 x 10^6 m/s, its kinetic energy is ½mv², or ½ (9.11 x 10^-31 kg) (9.70 x 10^6 m/s)² = 8.57 x 10^-17 J.

An electron passing through a 59.7 V potential gains 59.7 eV of energy (duh). Since 1 electron volt = 1.60217646 × 10^-19 joules, 59.7 eV = 9.552 x 10^-18 J

So the total energy of the electron after accelerating through the plates is 2.5257 x 10^-17 J. Since kinetic energy E = ½mv², solving for v gives

v = √[2E / m]

Substituting in the new energy and mass, we find the velocity to be

v = √[2 (2.5257x 10^-17) / (9.11 x 10^-31 kg)] = 1.4461x 10^7m/s

=14461 km/s