# Coulomb's Law

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Coulomb's law, named after Charles-Augustin Coulomb, is the fundamental law of electrostatic forces. It states that

The magnitude of the electrostatic force between two point charges is directly proportional to the magnitudes of each charge and inversely proportional to the square of the distance between the charges.

$F = \frac{q_1q_2}{4 \pi \epsilon\, r^2}$

Where:

$F \$ is the magnitude of the force exerted,
$q_1 \$ is the charge on one body,
$q_2 \$ is the charge on the other body,
$r \$ is the distance between them,
$\epsilon \$ is the permittivity of free space or permittivity of the vacuum. It is 8.854×10−12 C2 N-1 m-2 (also F m-1)
The occurrence of $4 \pi$ with the constant is related to some geometrical considerations that make some formulas (notably Maxwell's Equations) simpler. Ultimately it has to do with the fact that the surface area of a sphere is $4 \pi$. Much ink has been spilled over this issue—one sometimes sees equations in "rationalized SI units" instead of the "plain" SI units that we use here.

The force is in the direction of making like charges repel and unlike charges attract. It could be written in vector form as:

$\vec F_{2,1} = \frac{q_1q_2}{4 \pi \epsilon\, r^2} \hat r_{2,1}$

Where:

$F_{2,1}$ is the force vector acting on charge 2, coming from charge 1
$\hat r_{2,1}$ is the unit vector from particle 1 to particle 2, that is, the actual spatial vector divided by its length. Its length is of course $r$. The formula could be written as:
$\vec F_{2,1} = \frac{q_1q_2}{4 \pi \epsilon\, r^3} \vec r_{2,1}$
It is sometimes useful, in investigations of optics and capacitor dielectrics, to use a value of ε other than the standard "vacuum" value, since doing so can automatically handle the behavior of the dielectric. When this is done, it is common to use the symbol $\epsilon_0$ ("epsilon-nought") to denote the vacuum value, and $\epsilon$ to denote the modified value. When this is done, the "pure" formula is:
$F = \frac{q_1q_2}{4 \pi \epsilon_0\, r^2}$

## The Electric Force Field

Physicists usually reformulate this in terms of an electric field, usually denoted $\vec{E}$, that every charged object creates and that acts on every other charged object. Viewed this way, Coulomb's law breaks into two parts. The first part says what the field is around every charged object:

$\vec E = \frac{q}{4 \pi \epsilon\, r^2} \hat r_{2,1}$

where $r$ and $\hat r$ give the position at which the field is measured, and $q$ is the charge on the object creating the field.

The second parts is:

$\vec F = q \vec E$

where $q$ in this case is the charge of the particle being acted upon.

## The Electric Field Near a Very Long Uniformly Charged Wire

As we have seen, the electric field around a point charge is spherically symmetric and inversely proportional to the square of the distance. The are two other geometrical configurations worth looking at.

If we have an "infinitely long" linear collection of uniformly distributed charge (that is, a long charged wire), we can determine the nearby electric field by integration. Let the charge per unit length be $\rho$ coulombs per meter.

The electric field near a uniformly charged wire.

At a given point at distance $R$ from the wire, the contribution to the field from an infinitesimal section of wire of length dl is:

$\frac{\rho}{4 \pi \epsilon\, r^2}\ dl$

($\rho dl$ is the charge on that section of wire.)

The component of that vector pointing perpendicularly away from the wire is:

$\frac{\rho}{4 \pi \epsilon\, r^2}\ \sin\theta dl = \frac{\rho R}{4 \pi \epsilon\, r^3} dl = \frac{\rho R}{4 \pi \epsilon\, (R^2+l^2)^{3/2}} dl$

As we integrate over l from $-\infty$ to $+\infty$, the component of the field parallel to the wire cancels to zero, leaving:

$\frac{\rho R}{4 \pi \epsilon} \int_{-\infty}^\infty\frac{dl}{(R^2+l^2)^{3/2}} = \frac{\rho R}{4 \pi \epsilon} \frac{1}{R^2}\left.\frac{l}{\sqrt{R^2+l^2}}\right|_{-\infty}^\infty = \frac{\rho}{2 \pi \epsilon\, R}$

The field points perpendicularly away from the wire, and is inversely proportional to the first power of the separation distance.

Does the wire need to be infinitely long? No; this is just an approximation to the infinite limit. The approximation is good as long as one is much closer to the wire than the wire's length.

## The Electric Field Near a Very Large Uniformly Charged Plane

Another very important geometrical configuration is an "infinitely large" flat plane with uniform charge distribution. We divide the plane into many thin parallel strips of width dl. If the charge density per unit area of the plane is $\sigma$ coulombs per square meter, each strip has a linear charge density of

$\rho = \sigma dl\,$

coulombs per meter.

We use the result of the preceding section, and essentially the same diagram. The strips are now running into or out of the page / screen, and the strips' cross sections appear left-to-right on the diagram. The field at the point of interest is:

$\frac{\rho}{2 \pi \epsilon\, r} = \frac{\sigma}{2 \pi \epsilon\, r} dl$

The upward component (by symmetry the total field will point perpendicularly out of the plane) is:

$\frac{\sigma}{2 \pi \epsilon\, r}\ \sin\theta dl = \frac{\sigma R}{2 \pi \epsilon\, r^2}\ dl = \frac{\sigma R}{2 \pi \epsilon (R^2+l^2)}\ dl$

The total upward component of the field is obtained by integrating:

$E = \frac{\sigma R}{2 \pi \epsilon} \int_{-\infty}^\infty\frac{dl}{R^2+l^2} = \frac{\sigma}{2 \pi \epsilon} \left.\tan^{-1}\frac{l}{R}\right|_{-\infty}^\infty = \frac{\sigma}{2 \epsilon}$

The field points perpendicularly away from the plane, and is independent of the separation distance. That is, it is independent as long as one stays so close to the plane that it appears to be nearly infinite.