Question

Two electrons are separated by 1.70 nm. What is the magnitude of the electric force each electron exerts on the other?

Answer 1

`F=7.96*10^(-11)N`

Step-by-step explanation:

According to Coulomb's law, the magnitude of the electric force between two equals charges (q) is given by:

`F=(kq^2)/(d^2)`

Here k is the coulomb constant and d is the distance between the charges. For two electrons we have:

`F=(ke^2)/(d^2)\\F=(8.99*10^(9)(N\cdot m^2)/(C^2)(-1.6*10^(-19)C)^2)/((1.7*10^(-9)m)^2)\\F=7.96*10^(-11)N`