An initially stationary electron is accelerated by a uniform 640 N/C Electric Field. a) Calculate the kinetic energy of the electron after it has traveled 15 cm in a direction parallel to this field. b) Calculate the speed of the electron after it has traveled 15 cm in a direction parallel to this field.

(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.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west