Answer:
(a) 1.298 * 10^(-4) J
(b) 5.82 * 10^6 m/s
Explanation:
Parameters given:
Electric field, E = 640 N/C
Distance traveled by electron, r = 15 cm = 0.15 m
Mass of electron, m = 9.11 * 10^(-31) kg
Electric charge of electron, q = 1.602 * 10^(-19) C
(a) The kinetic energy of the electron in terms of Electric field is given as:
K = (q² * E² * r²) / 2m
Therefore, Kinetic energy, K, is:
K = [(1.602 * 10^(-19))² * 640² * 0.15²] / [2 * 9.11 * 10^(-31)]
K = {23651.981 * 10^(-38)} / [18.22 * 10^(-31)]
K = 1298.13 * 10^(-7) J = 1.298 * 10^(-4) J
(b) To find the final velocity of the electron, we have to first find the acceleration of the electron. This can be gotten by using the equations of force.
Force is generally given as:
F = ma
Electric force is given as:
F = qE
Therefore, equating both, we have:
ma = qE
a = (qE) / m
a = (1.602 * 10^(-19) * 640) / (9.11 * 10^(-31))
a = 112.54 * 10^(12) m/s² = 1.13 * 10^(14) m/s²
Using one of the equations of motion, we have that:
v² = u² + 2as
Since the electron started from rest, u = 0 m/s
Therefore:
v² = 2 * 1.13 * 10^(14) * 0.15
v² = 3.39 * 10^(13)
v = 5.82 * 10^6 m/s
The velocity of the electron after moving a distance of 15 cm is 5.82 * 10^6 m/s.
