electron question?

Question

What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 N/C and is directed due north?

Answer 1

force on electron = e x E
e---------charge on electron = 1.6 x 10 ^ (-19) C
E---------electric field applied = 6100N/C
F = e X E
m X a = e X E
m------mass of electron=9.1 x 10^(-31) kg
a ----- acceleration of electron

a = e x E/m
a = 1.6 x 10^(-19) x 6100 / 9.1 x 10^(-31)
a = 16 x 6100 x 10^(12) / 91
a = 1.072527 x 10^(15) m/s2 ( -j) , i.e., in south

Answer 2

The net force (F) on the electron will be due to the electric field (E).
F = q * E
where q is the charge on the electron.
|q| = 1.6 E-19 Coulombs

F = 9.76 E-16 Newtons

The acceleration is always in the direction of the net force.
Force = mass * acceleration
a = Force / mass

The mass of an electron is: 9.11 E-31 kg

a = 9.76 E-16 Newtons / 9.11 E-31 kg
a = 1.07 E15 m/s^2

Electrons (since they are negatively charged) will want to move with the electric field, so the net force will be pointing North, as will the acceleration.

The acceleration experienced by the electron will have a magnitude of 1.07 E15 m/s^2 and will be directed due north along the electric field.