Explanation:
We have,
Electric field intensity is 700 N/Cj
It is required to find the magnitude and direction of the acceleration of this electron due to this field.
The electric force is balanced by the force due to its motion i.e.
qE =ma
a is acceleration of this electron
[tex]a=\dfrac{qE}{m}\\\\a=\dfrac{1.6\times 10^{-19}\times 700}{9.1\times 10^{-31}}\\\\a=1.23\times 10^{14}\ m/s^2[/tex]
So, the acceleration of the electron is [tex]1.23\times 10^{14}\ m/s^2[/tex] and it is moving in +y direction.
