Question

An electron enters a region of uniform electric field with an initial velocity of 40 km/s in the same direction as the electric field, which has magnitude E=50 N/C.

What is the speed of the electron 1.5 ns after entering this region?

Answer 1

Vf = 53186.8 m/s = 53.18 km/s

Step-by-step explanation:

First we calculate the force applied on the electron by the field, by using the following equation:

E = F/q

F = E q

where,

F = Force = ?

E = Electric Field = 50 N/C

q = Charge on Electron = 1.6 x 10⁻¹⁹ C

Therefore,

F = (50 N/C)(1.6 x 10⁻¹⁹ C)

F = 8 x 10⁻¹⁸ N

Now, we calculate the acceleration of electron:

F = ma

a = F/m

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

a = acceleration = ?

Therefore,

a = (8 x 10⁻¹⁸ N)/(9.1 x 10⁻³¹ kg)

a = 0.88 x 10¹³ m/s²

Now, we use 1st equation of motion:

Vf = Vi + at

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 40 km/s = 40000 m/s

t = time taken = 1.5 ns = 1.5 x 10⁻⁹ s

Therefore,

Vf = 40000 m/s + (0.88 x 10¹³ m/s²)(1.5 x 10⁻⁹ s)

Vf = 53186.8 m/s = 53.18 km/s