Given,
The electric force experienced by the electron, F=6.4×10⁻¹⁴ N
The charge of an electron, q_e=1.6×10⁻¹⁹ C
The electric field is the physical field in which if a charged particle is placed, it experiences the electric force.
The electric field and electric force are related to each other using the formula,
[tex]F=Eq[/tex]
On substituting the known values,
[tex]\begin{gathered} 6.4\times10^{-14}=E\times1.6\times10^{-19} \\ E=\frac{6.4\times10^{-14}}{1.6\times10^{-19}} \\ =400\times10^3\text{ N/C} \end{gathered}[/tex]
Therefore the electric field at that location is 400×10³ N/C
