# What is Coulomb's law

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

Coulomb's law is an inverse-square law stating that the force vector between two stationary charges is a constant times the unit vector between them and times the product of the magnitudes of the charges divided by the square of the distance between them: $\mathbf{F}_{12}\ \propto\ Q_1Q_2\,\mathbf{\hat{r}}_{12}/r_{12}^2$

That constant (Coulomb's constant) is the same in any material, and is $1/4\,\pi\,\varepsilon_0$, where $\varepsilon_0$ is the electric constant (the permittivity of the vacuum, with dimensions of charge²/force.area, and measured in units of farads/metre), and $4\,\pi$ is the ratio between the surface area of a sphere and the radius squared.

If the charges have the same sign, then $Q_1Q_2$ is positive, and the force vector points outward (the force is repulsive); if they have opposite signs, then $Q_1Q_2$ is negative, and the force vector points inward (the force is attractive).

Gauss' law (one of Maxwell's equations) may be derived from Coulomb's law.

Equations

Force on stationary charge 2 from stationary charge 1:

$$\mathbf{F}_{12}\ =\ \frac{Q_1\,Q_2}{4\,\pi\,\varepsilon_0\,r_{12}^2}\,\mathbf{\hat{r}}_{12}$$

Electric field of charge 1 at position of charge 2 (from Lorentz force equation):

$$\mathbf{E}_{12}\ =\ \frac{\mathbf{F}_{12}}{Q_2}\ =\ \frac{Q_1}{4\,\pi\,\varepsilon_0\,r_{12}^2}\,\mathbf{\hat{r}}_{12}$$

Since this is independent of the magnitude of charge 2, it may be rewritten:

$$\mathbf{E}_1(\mathbf{r})\ =\ \frac{Q_1}{4\,\pi\,\varepsilon_0\,r^2}\,\mathbf{\hat{r}}\ =\ \frac{Q_1}{\varepsilon_0\,A(r)}\,\mathbf{\hat{r}}$$

where $A(r)$ is the surface area of the sphere $S(r)$ through $\mathbf{r}$ with charge 1 at its centre.

Obviously, the divergence of $\mathbf{E}_1$ at any point other than the position of charge 1 is zero (differential form of Gauss' law for zero charge density $\rho$):

$$\nabla\cdot\mathbf{E}_1\ =\ 0\ \ \text{if}\ \ \rho\ =\ 0$$

And the flux of $\mathbf{E}_1$ through the sphere $S(r)$ is $Q_1/\varepsilon_0$:

$$\oint_{S(r)}\,\mathbf{E}_1\cdot(\mathbf{\hat{r}}\,dA)\ =\ \frac{Q_1}{\varepsilon_0 \,A(r)}\,\oint_{S(r)}\,\mathbf{\hat{r}}\cdot\mathbf{\hat{r}}\,dA\ =\ \frac{Q_1}{\varepsilon_0 \,A(r)}\,\oint_{S(r)}dA\ =\ \frac{Q_1}{\varepsilon_0}$$

and so, from Stoke's theorem, the flux of $\mathbf{E}_1$ through any closed surface S containing charge 1 is $Q_1/\varepsilon_0$ and through any other closed surface is zero:

$$\oint_S \, \mathbf{E}_1 \cdot(\mathbf{\hat{n}} \, dA)\ =\ \left\{\begin{array}{cc} Q_1/\varepsilon_0 & \text{if S contains charge 1}\\ 0 & \text{if S does not contain charge 1}\end{array}\right.$$

Extended explanation

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