Question

An isolated electron experiences an electric force of 6.4 x 10-14 N. What is the magnitude of the electric field at that location? (qe = 1.6 x 10-19 C)Question 3

Answer 1

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,

`F=Eq`

On substituting the known values,

`\begin{gathered} 6.4*10^(-14)=E*1.6*10^(-19) \\ E=(6.4*10^(-14))/(1.6*10^(-19)) \\ =400*10^3\text{ N/C} \end{gathered}`

Therefore the electric field at that location is 400×10³ N/C