Student Question

# An electron with velocity v = 2.0x10-6 m/s directed along +y axis enters in a region of uniform electric field, E = 8.0 x 103 V/m directed along +y axis. What is the magnitude and direction of the acceleration of the electron? How long does it take for the electron to stop?

When the electron enter into the electric field, the electric force acts on it. In this case the electric force is calculated using the following expression:

Fe = -e*E

Where -e is the charge of the electron and E is the electric field strength.

Applying Newton's second law, we can calculate the acceleration experienced by the electron:

F = m*a

In our case, F is the electric force and m is the mass of the electron, i.e.:

Fe = me*a

a = Fe/me = (-e*E)/me

a = (-1.602*10^-19)(8.0*10^3)/(9.109*10^-31)

a = -1.4*10^15 m/s^2

As the charge of the electron is negative, the direction of the force is contrary to the field, that is, along the -y axis.

To find the time that it takes to stop, we apply the equation of acceleration:

a = (v – v0)/t

Where v0 is the initial speed and v is the final velocity, which in this case is zero.

t = (v – v0)/a = (0 – 2.0*10^-6)/(- 1.4*10^15)

t = 1.43*10^-21 s