# The Unapologetic Mathematician

## The Electric Field

What happens if instead of two particles we have three? For simplicity, let’s just consider the resultant force on one of the three particles; say it has charge $q$ and the other particles have charges $q_1$ and $q_2$, with displacement vectors $r_1$ and $r_2$, respectively. The Coulomb law tells us that the first of the other particles exerts a force

$\displaystyle F_1=\frac{1}{4\pi\epsilon_0}\frac{qq_1}{\lvert r_1\rvert^3}r_1$

while the second exerts a force

$\displaystyle F_2=\frac{1}{4\pi\epsilon_0}\frac{qq_2}{\lvert r_2\rvert^3}r_2$

As usual, we just add the forces together to get the resultant

\displaystyle\begin{aligned}F&=F_1+F_2\\&=\frac{1}{4\pi\epsilon_0}\frac{qq_1}{\lvert r_1\rvert^3}r_1+\frac{1}{4\pi\epsilon_0}\frac{qq_2}{\lvert r_2\rvert^3}r_2\\&=q\left(\frac{1}{4\pi\epsilon_0}\frac{q_1}{\lvert r_1\rvert^3}r_1+\frac{1}{4\pi\epsilon_0}\frac{q_2}{\lvert r_2\rvert^3}r_2\right)\end{aligned}

Of course, as we add more particles we just add more terms to the sum. But always we find that the force on the “test” particle with charge $q$ times some vector field: $F=qE$. Coulomb’s law tells us that a single point of charge $q'$ at the origin gives rise to a vector field whose value at the point with position vector $r$ is

$\displaystyle E=\frac{1}{4\pi\epsilon_0}\frac{q'}{\lvert r\rvert^3}r$

That is, it’s sort of like the radial vector field only instead of getting larger as we move away from the origin, it gets smaller, falling off as the square of the distance.

As we’ve just seen, the superposition principle for forces leads to a superposition principle for the electric field: the field generated by two or more sources is the (vector) sum of the fields generated by each source separately.

January 5, 2012