Lecture 1.1 - Columbs Law

ELEC2300

Coulombs Law

$$F=k\frac{q_1{q}_2}{r_{12}^2}$$

We need to remember that

$$k=\frac{1}{4\pi {ϵ}_0}=8.99\times 10^9$$

Coulomb’s Law only applies to point charges in a field (whether it be empty or an electric field).

In this case, the force applied in the equation is a scalar, not a vector. To redefine into a vector:

$${\vec{F}}_{12}={\frac{k{q}_1{q}_2}{r^2}}_{12}\times {\hat{r}}_{12}$$