Coulomb's law

When the electromagnetic theory is expressed in the International System of Units, force is measured in newtons, charge in coulombs, and distance in meters. Coulomb's constant is given by k_e = 1/4πε₀. The constant ε₀ is the vacuum electric permittivity (also known as "electric constant") [23] in C²⋅m⁻²⋅N⁻¹. It should not be confused with ε_r, which is the dimensionless relative permittivity of the material in which the charges are immersed, or with their product ε_a = ε₀ε_r , which is called "absolute permittivity of the material" and is still used in electrical engineering.

The SI derived units for the electric field are volts per meter, newtons per coulomb, or tesla meters per second.

Coulomb's law and Coulomb's constant can also be interpreted in various terms:

• Atomic units. In atomic units the force is expressed in hartrees per Bohr radius, the charge in terms of the elementary charge, and the distances in terms of the Bohr radius.
• Electrostatic units or Gaussian units. In electrostatic units and Gaussian units, the unit charge (esu or statcoulomb) is defined in such a way that the Coulomb constant k disappears because it has the value of one and becomes dimensionless.
• Lorentz–Heaviside units (also called rationalized). In Lorentz–Heaviside units the Coulomb constant is k_e = 1/4π and becomes dimensionless.

Gaussian units and Lorentz–Heaviside units are both CGS unit systems. Gaussian units are more amenable for microscopic problems such as the electrodynamics of individual electrically charged particles.[24] SI units are more convenient for practical, large-scale phenomena, such as engineering applications.[24]

If two charges have the same sign, the electrostatic force between them is repulsive; if they have different sign, the force between them is attractive.

An electric field is a vector field that associates to each point in space the Coulomb force experienced by a test charge. In the simplest case, the field is considered to be generated solely by a single source point charge. The strength and direction of the Coulomb force F on a test charge q_t depends on the electric field E that it finds itself in, such that F = q_tE. If the field is generated by a positive source point charge q, the direction of the electric field points along lines directed radially outwards from it, i.e. in the direction that a positive point test charge q_t would move if placed in the field. For a negative point source charge, the direction is radially inwards.

The magnitude of the electric field E can be derived from Coulomb's law. By choosing one of the point charges to be the source, and the other to be the test charge, it follows from Coulomb's law that the magnitude of the electric field E created by a single source point charge q at a certain distance from it r in vacuum is given by:

${\displaystyle |{\boldsymbol {E}}|={1 \over 4\pi \varepsilon _{0}}{|q| \over r^{2}}}$

Coulomb's constant is a proportionality factor that appears in Coulomb's law as well as in other electric-related formulas. The value of this constant is dependent upon the medium that the charged objects are immersed in. Denoted k_e, it is also called the electric force constant or electrostatic constant,[8] hence the subscript e.

Prior to the 2019 redefinition of the SI base units, the Coulomb constant (in the case of air or vacuum) was considered to have an exact value:

${\displaystyle {\begin{aligned}k_{e}&={\frac {1}{4\pi \varepsilon _{0}}}={\frac {c_{0}^{2}\mu _{0}}{4\pi }}=c_{0}^{2}\times 10^{-7}\ \mathrm {H\cdot m} ^{-1}\\&=8.987\,551\,787\,368\,176\,4\times 10^{9}\ \mathrm {N\cdot m^{2}\cdot C} ^{-2}.\end{aligned}}}$

Since the 2019 redefinition,[25][26] the Coulomb constant is no longer exactly defined and is subject to the measurement error in the fine structure constant. As calculated from CODATA 2018 recommended values, the Coulomb constant is[27]

${\displaystyle k_{\text{e}}=8.987\,551\,792\,3\,(14)\times 10^{9}\ \mathrm {N\cdot m^{2}\cdot C} ^{-2}.}$

There are three conditions to be fulfilled for the validity of Coulomb's law:

1. The charges must have a spherically symmetric distribution (e.g. be point charges, or a charged metal sphere).
2. The charges must not overlap (e.g. they must be distinct point charges).
3. The charges must be stationary with respect to each other.

The last of these is known as the electrostatic approximation. When movement takes place, Einstein's theory of relativity must be taken into consideration, and a result, an extra factor is introduced, which alters the force produced on the two objects. This extra part of the force is called the magnetic force, and is described by magnetic fields. For slow movement, the magnetic force is minimal and Coulomb's law can still be considered approximately correct, but when the charges are moving more quickly in relation to each other, the full electrodynamic rules (incorporating the magnetic force) must be considered.

The most basic Feynman diagram for QED interaction between two fermions

In simple terms, the Coulomb potential derives from the QED Lagrangian as follows. The Lagrangian of quantum electrodynamics is normally written in natural units, but in SI units, it is:

${\displaystyle {\mathcal {L}}_{\mathrm {QED} }={\bar {\psi }}(i\hbar c\gamma ^{\mu }D_{\mu }-mc^{2})\psi -{\frac {1}{4\hbar c}}F_{\mu \nu }F^{\mu \nu }}$

where the covariant derivative (in SI units) is:

${\displaystyle {D}_{\mu }=\partial _{\mu }+{\frac {ig}{\hbar c}}A_{\mu }}$

where ${\displaystyle g}$ is the gauge coupling parameter. By putting the covariant derivative into the lagrangian explicitly, the interaction term (the term involving both ${\displaystyle A}$ and ${\displaystyle \psi }$) is seen to be:

${\displaystyle {\mathcal {L}}_{\mathrm {int} }=ig{\bar {\psi }}\gamma ^{\mu }A_{\mu }\psi }$

The most basic Feynman diagram for a QED interaction between two fermions is the exchange of a single photon, with no loops. Following the Feynman rules, this therefore contributes two QED vertex factors (${\displaystyle igQ\gamma _{\mu }}$) to the potential, where Q is the QED-charge operator (Q gives the charge in terms of the electron charge, and hence is exactly −1 for electrons, etc.). For the photon in the diagram, the Feynman rules demand the contribution of one bosonic massless propagator ${\displaystyle \left({\frac {\hbar c}{k^{2}}}\right)}$. Ignoring the momentum on the external legs (the fermions), the potential is therefore:

${\displaystyle V(\mathbf {r} )={\frac {1}{(2\pi )^{3}}}\int e^{\frac {i\mathbf {k\cdot r} }{\hbar }}(iQ_{1}g^{2}\gamma _{\mu })(iQ_{2}g^{2}\gamma _{\nu })\left({\frac {\hbar c}{k^{2}}}\right)\;d^{3}k}$

which can be more usefully written as

${\displaystyle V(\mathbf {r} )={\frac {-g^{2}\hbar cQ_{1}Q_{2}}{4\pi }}{\frac {1}{(2\pi )^{3}}}\int e^{\frac {i\mathbf {k\cdot r} }{\hbar }}{\frac {4\pi \eta _{\mu \nu }}{k^{2}}}\;d^{3}k}$

where ${\displaystyle Q_{i}}$ is the QED-charge on the ith particle. Recognising the integral as just being a Fourier transform enables the equation to be simplified:

${\displaystyle V(\mathbf {r} )={\frac {-g^{2}\hbar cQ_{1}Q_{2}}{4\pi }}{\frac {1}{r}}.}$

For various reasons, it is more convenient to define the fine-structure constant ${\displaystyle \alpha ={\frac {g^{2}}{4\pi }}}$, and then define ${\displaystyle e={\sqrt {4\pi \alpha \varepsilon _{0}\hbar c}}}$. Rearranging these definitions gives:

${\displaystyle g^{2}\hbar c={\frac {e^{2}}{\varepsilon _{0}}}}$

In natural units (${\displaystyle \hbar =1}$, ${\displaystyle c=1}$, and ${\displaystyle \varepsilon _{0}=1}$), ${\displaystyle g=e}$. Continuing in SI units, the potential is therefore

${\displaystyle V(\mathbf {r} )={\frac {-e^{2}}{4\pi \varepsilon _{0}}}{\frac {Q_{1}Q_{2}}{r}}}$

Defining ${\displaystyle q_{i}=eQ_{i}}$, as the macroscopic 'electric charge', makes e the macroscopic 'electric charge' for an electron, and enables the formula to be put into the familiar form of the Coulomb potential:

${\displaystyle V(\mathbf {r} )={\frac {-1}{4\pi \varepsilon _{0}}}{\frac {q_{1}q_{2}}{r}}}$

The force (${\displaystyle {\frac {dV(\mathbf {r} )}{dr}}}$) is therefore :

${\displaystyle F(\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}{\frac {q_{1}q_{2}}{r^{2}}}}$

The derivation makes clear that the force law is only an approximation — it ignores the momentum of the input and output fermion lines, and ignores all quantum corrections (i.e. the myriad possible diagrams with internal loops).

The Coulomb potential, and its derivation, can be seen as a special case of the Yukawa potential (specifically, the case where the exchanged boson – the photon – has no rest mass).

The law of superposition allows Coulomb's law to be extended to include any number of point charges. The force acting on a point charge due to a system of point charges is simply the vector addition of the individual forces acting alone on that point charge due to each one of the charges. The resulting force vector is parallel to the electric field vector at that point, with that point charge removed.

The force F on a small charge q at position r, due to a system of N discrete charges in vacuum is:

${\displaystyle {\boldsymbol {F(r)}}={q \over 4\pi \varepsilon _{0}}\sum _{i=1}^{N}q_{i}{{\boldsymbol {r-r_{i}}} \over |{\boldsymbol {r-r_{i}}}|^{3}}={q \over 4\pi \varepsilon _{0}}\sum _{i=1}^{N}q_{i}{{\boldsymbol {\widehat {R_{i}}}} \over |{\boldsymbol {R_{i}}}|^{2}},}$

where q_i and r_i are the magnitude and position respectively of the ith charge, R̂_i is a unit vector in the direction of R_i = r − r_i (a vector pointing from charges q_i to q).[29]

In this case, the principle of linear superposition is also used. For a continuous charge distribution, an integral over the region containing the charge is equivalent to an infinite summation, treating each infinitesimal element of space as a point charge dq. The distribution of charge is usually linear, surface or volumetric.

For a linear charge distribution (a good approximation for charge in a wire) where λ(r′) gives the charge per unit length at position r′, and dℓ′ is an infinitesimal element of length,

${\displaystyle dq=\lambda ({\boldsymbol {r'}})\,d\ell '.}$[30]

For a surface charge distribution (a good approximation for charge on a plate in a parallel plate capacitor) where σ(r′) gives the charge per unit area at position r′, and dA′ is an infinitesimal element of area,

${\displaystyle dq=\sigma ({\boldsymbol {r'}})\,dA'.}$

For a volume charge distribution (such as charge within a bulk metal) where ρ(r′) gives the charge per unit volume at position r′, and dV′ is an infinitesimal element of volume,

${\displaystyle dq=\rho ({\boldsymbol {r'}})\,dV'.}$[29]

The force on a small test charge q′ at position r in vacuum is given by the integral over the distribution of charge:

${\displaystyle {\boldsymbol {F}}={q' \over 4\pi \varepsilon _{0}}\int dq{{\boldsymbol {r}}-{\boldsymbol {r'}} \over |{\boldsymbol {r}}-{\boldsymbol {r'}}|^{3}}.}$

Strictly speaking, Gauss's law cannot be derived from Coulomb's law alone, since Coulomb's law gives the electric field due to an individual point charge only. However, Gauss's law can be proven from Coulomb's law if it is assumed, in addition, that the electric field obeys the superposition principle. The superposition principle says that the resulting field is the vector sum of fields generated by each particle (or the integral, if the charges are distributed smoothly in space).

Outline of proof

Coulomb's law states that the electric field due to a stationary point charge is: ${\displaystyle \mathbf {E} (\mathbf {r} )={\frac {q}{4\pi \varepsilon _{0}}}{\frac {\mathbf {e} _{r}}{r^{2}}}}$ where er is the radial unit vector, r is the radius, |r|, ε0 is the electric constant, q is the charge of the particle, which is assumed to be located at the origin. Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give ${\displaystyle \mathbf {E} (\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\int {\frac {\rho (\mathbf {s} )(\mathbf {r} -\mathbf {s} )}{|\mathbf {r} -\mathbf {s} |^{3}}}\,\mathrm {d} ^{3}\mathbf {s} }$ where ρ is the charge density. If we take the divergence of both sides of this equation with respect to r, and use the known theorem[31] ${\displaystyle \nabla \cdot \left({\frac {\mathbf {r} }{|\mathbf {r} |^{3}}}\right)=4\pi \delta (\mathbf {r} )}$ where δ(r) is the Dirac delta function, the result is ${\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r} )={\frac {1}{\varepsilon _{0}}}\int \rho (\mathbf {s} )\,\delta (\mathbf {r} -\mathbf {s} )\,\mathrm {d} ^{3}\mathbf {s} }$ Using the "sifting property" of the Dirac delta function, we arrive at ${\displaystyle \nabla \cdot \mathbf {E} (\mathbf {r} )={\frac {\rho (\mathbf {r} )}{\varepsilon _{0}}},}$ which is the differential form of Gauss' law, as desired.

Note that since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more general than Coulomb's law.

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

Outline of proof

Taking S in the integral form of Gauss' law to be a spherical surface of radius r, centered at the point charge Q, we have ${\displaystyle \oint _{S}\mathbf {E} \cdot d\mathbf {A} ={\frac {Q}{\varepsilon _{0}}}}$ By the assumption of spherical symmetry, the integrand is a constant which can be taken out of the integral. The result is ${\displaystyle 4\pi r^{2}{\hat {\mathbf {r} }}\cdot \mathbf {E} (\mathbf {r} )={\frac {Q}{\varepsilon _{0}}}}$ where r̂ is a unit vector pointing radially away from the charge. Again by spherical symmetry, E points in the radial direction, and so we get ${\displaystyle \mathbf {E} (\mathbf {r} )={\frac {Q}{4\pi \varepsilon _{0}}}{\frac {\hat {\mathbf {r} }}{r^{2}}}}$ which is essentially equivalent to Coulomb's law. Thus the inverse-square law dependence of the electric field in Coulomb's law follows from Gauss' law.