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.

(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.

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-25T12:34:05+01:00

(a) The kinetic energy of the electron is 1.53×10⁻¹⁷ J

(b) The final speed of the electron is 5.8×10⁶ m/s

Kinetic energy and final speed:

The electron is under motion due to the electrostatic force F produced by the electric field E = 640 N/C.

The equation of motion of the electron can be written as:

F = qE

⇒ ma = qE

where q = 1.6×10⁻¹⁹ C is the charge on the electron

m = 9.1×10⁻³¹ kg is the mass of the electron

and a is the acceleration.

So,

a = qE/m

a = 1.6×10⁻¹⁹×640 / 9.1×10⁻³¹

a = 1.12×10¹⁴ m/s²

The initial speed of the electron is zero, so u = 0

From the third equation of motion:

v² = u² + 2as

where v is the final speed,

and s is the distance traveled = 15cm = 0.15m

v² = 0 + 2× 1.12×10¹⁴×0.15

v = 5.8×10⁶ m/s

So the Kinetic energy of the electron is:

KE = ¹/₂mv²

KE = 0.5×9.1×10⁻³¹×(5.8×10⁶)²

KE = 1.53×10⁻¹⁷ J

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