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

Answer:

$F=7.96*10^{-11}N$

Explanation:

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

$F=\frac{kq^2}{d^2}$

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

$F=\frac{ke^2}{d^2}\\F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)^2}{(1.7*10^{-9}m)^2}\\F=7.96*10^{-11}N$