Question

An electron is released from rest in a uniform electric of magnitude 2.00×104 N/C. Calculate the acceleration of the electron (Ignore gravitation)?

Answer 1

`a = 3.51 xx 10^15 m/s^2`
This would be magnitude of the electron’s acceleration. Since the electron has a negative charge the direction of the force on the electron and of course the acceleration is opposite the direction of the electric field.

Explanation:

calculate Force first
Force `F = q_eE`, where `q_e= 1.602xx10^-19 C` charge of electron
`E = 2xx10^4 N/C` electric field
now divide force by mass of electron to have the acceleration.
`a = F/m`, where `m= 9.11xx10^-31` mass of electron
`a = q_eE/m = (1.602xx10^-19 C xx 2.0xx10^4 N/C)/(9.11xx10^-31 kg)`
`a = 3.51 xx 10^15 m/s^2`
This would be magnitude of the electron’s acceleration. Since the electron has a negative charge the direction of the force on the electron and of course the acceleration is opposite the direction of the electric field.