Coulomb's law

In physics, Coulomb's law describes the forces acting between electric point charges. The law was first given by Charles-Augustin de Coulomb. It is an inverse-square law for two electric charges very similar to Newton's gravitational law for two masses. An important difference between the two laws is that masses always attract each other, whereas charges may repel or attract. That is, charges may be positive or negative, while masses have the same sign (are always positive by convention).

Formulation

Coulomb discovered experimentally that the force between two small charged bodies separated in air a distance large compared to their dimensions

1. varies directly as the magnitude of each charge,
2. varies inversely as the square of the distance between them,
3. is directed along the line joining the charges,
4. is attractive if the bodies are oppositely charged and repulsive if the bodies have the same type of charge.

Further it was shown experimentally that the force is additive, i.e., the force on a test body exerted by a number of charges around it, is the vector sum of the individual two-body forces. Given two charges q and q' a distance r apart, the force F between the particles is,

$F={\frac {q\,q'}{4\pi \epsilon _{0}\epsilon _{r}r^{2}}}$

The quantities ε₀ and ε_r are the vacuum permittivity and the relative static permittivity (also known as relative dielectric constant) of the medium, respectively. The relative dielectric constant of air is: ε_r = 1.000536 (at ambient temperature and pressure). The formulation of Coulomb's law is in the rationalized SI system of units.

Electrostatic vector field

Coulomb's law. Electric field E at point P due to positive charge q.

The force $\scriptstyle {\vec {\mathbf {F} }}$ divided by Δq on a test particle of charge Δq is the electrostatic field $\scriptstyle {\vec {\mathbf {E} }}$—a vector field. The force originates from one or more charges acting on the test particle. The charge Δq of the test particle is taken infinitesimally small, that is, it is negligible with respect to the charges causing the field and hence the test particle does not influence the electric field. The direction of the electric field is by convention such that it points away from a positive charge and points toward a negative charge.

Coulomb's law gives the electric field $\scriptstyle {\vec {\mathbf {E} }}$ at the point P due to an electrostatic point charge q at position $\scriptstyle {\vec {\mathbf {r} }}$. The strength of the electric field $\scriptstyle {\vec {\mathbf {E} }}$ depends on the inverse-squared distance of P to q and is along the line from q to P, that is,

${\vec {\mathbf {E} }}={\frac {q({\vec {\mathbf {R} }}-{\vec {\mathbf {r} }})}{4\pi \epsilon _{0}\epsilon _{r}|{\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}|^{3}}}.$

Note that although this may look like an inverse-cubed dependence, we must not forget that the denominator has dimension length. Indeed, defining a (dimensionless) unit vector by

${\hat {\mathbf {r} }}_{qp}\equiv {\frac {{\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}}{|{\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}|}},$

we see more clearly the inverse-squared dependence on distance:

${\vec {\mathbf {E} }}={\frac {q{\hat {\mathbf {r} }}_{qp}}{4\pi \epsilon _{0}\epsilon _{r}|{\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}|^{2}}}.$

If q is positive the electric field points from q to P, if q is negative it points from P to q. The force on a particle of charge −|Q| is equal to $\scriptstyle -|Q|{\vec {\mathbf {E} }}$, that is, the force is antiparallel to the vector $\scriptstyle {\vec {\mathbf {E} }}$. The force on a particle with charge |Q| is equal to $\scriptstyle |Q|{\vec {\mathbf {E} }}$, parallel to the vector $\scriptstyle {\vec {\mathbf {E} }}$. In total, charges of opposite sign attract and of same sign repel.

The experimentally observed additivity of electrostatic forces gives the following electric field at P when there are N charges q_k at fixed positions r_k. The electric field is a vector field

${\vec {\mathbf {E} }}({\vec {\mathbf {R} }})=\sum _{k=1}^{N}{\frac {q_{k}({\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}_{k})}{4\pi \epsilon _{0}\epsilon _{r}|{\vec {\mathbf {R} }}-{\vec {\mathbf {r} }}_{k}|^{3}}}$

depending on $\scriptstyle {\vec {\mathbf {R} }}$. The sum stands for a vector sum.

Coulomb potential

From here on we denote vectors by bold letters dropping the arrows on top of the symbols.

When the electrostatic field is irrotational, i.e.,

${\boldsymbol {\nabla }}\times \mathbf {E} =\mathbf {0}$

one can define a potential Φ(r), such that,

$\mathbf {E} (\mathbf {R} )=-{\boldsymbol {\nabla }}\Phi (\mathbf {R} ).$

Since

${\frac {\partial }{\partial Y}}{\frac {Z}{R^{3}}}={\frac {\partial }{\partial Z}}{\frac {Y}{R^{3}}},\quad {\frac {\partial }{\partial X}}{\frac {Z}{R^{3}}}={\frac {\partial }{\partial Z}}{\frac {X}{R^{3}}},\quad {\frac {\partial }{\partial Y}}{\frac {X}{R^{3}}}={\frac {\partial }{\partial X}}{\frac {Y}{R^{3}}},$

it follows that R/R³ is irrotational, and so is (R-r)/|R-r|³. The Coulomb potential Φ exists. The following expression for Φ is consistent with E,

$\Phi (\mathbf {R} )={\frac {q}{4\pi \epsilon _{0}\epsilon _{r}|\mathbf {R} -\mathbf {r} |}}\quad \Longrightarrow \quad \mathbf {E} (\mathbf {R} )=-{\boldsymbol {\nabla }}{\frac {q}{4\pi \epsilon _{0}\epsilon _{r}|\mathbf {R} -\mathbf {r} |}}={\frac {q(\mathbf {R} -\mathbf {r} )}{4\pi \epsilon _{0}\epsilon _{r}|\mathbf {R} -\mathbf {r} |^{3}}}.$

The Coulomb potential is determined up to a constant. By requiring that Φ vanishes for infinite R the constant becomes zero.

The Coulomb potential at the point R due to N charges at r_k is

$\Phi (\mathbf {R} )=\sum _{k=1}^{N}{\frac {q_{k}}{4\pi \epsilon _{0}\epsilon _{r}|\mathbf {R} -\mathbf {r} _{k}|}}.$

For a continuous charge distribution ρ(r) this expression generalizes to

$\Phi (\mathbf {R} )=\iiint {\frac {\rho (\mathbf {r} )}{4\pi \epsilon _{0}\epsilon _{r}|\mathbf {R} -\mathbf {r} |}}d\mathbf {r} .$

This generalization follows easily if we first substitute the infinitesimal charge at position r,

$\Delta q(\mathbf {r} )=\rho (\mathbf {r} )d\mathbf {r} \equiv \rho (\mathbf {r} )dxdydz,$

and then replace the sum by a volume integral.

Poisson equation

The Maxwell equations form the postulates of classical electrodynamics. It is of interest to see how the Coulomb potential fits in the postulatory framework devised by James Clerk Maxwell. The following two Maxwell equations are of relevance in the present case in which only static electric charges are considered,

${\boldsymbol {\nabla }}\cdot \mathbf {E} =\rho /\epsilon _{0}\quad {\hbox{and}}\quad {\boldsymbol {\nabla }}\times \mathbf {E} =\mathbf {0} .$

Here we assumed ε_r = 1 (the vacuum case). The second Maxwell equation tells us that there exists a potential Φ with

$\mathbf {E} =-{\boldsymbol {\nabla }}\Phi ,$

so that we obtain

${\boldsymbol {\nabla }}\cdot {\boldsymbol {\nabla }}\Phi \equiv \nabla ^{2}\Phi =-\rho /\epsilon _{0}.$

This equation, which predates Maxwell's equations, was postulated by Siméon Denis Poisson.

In order to show that the Coulomb potential, introduced above, satisfies the Poisson equation, we first recall the following result from distribution theory,

$\nabla ^{2}\left({\frac {1}{|\mathbf {R} -\mathbf {r} |}}\right)=-4\pi \delta (\mathbf {R} -\mathbf {r} ),$

where the right hand side is the Dirac delta function. Now,

$\nabla ^{2}\Phi (\mathbf {R} )={\frac {1}{4\pi \epsilon _{0}}}\iiint \rho (\mathbf {r} )\nabla ^{2}{\Big (}{\frac {1}{|\mathbf {R} -\mathbf {r} |}}{\Big )}d\mathbf {r} =-{\frac {1}{\epsilon _{0}}}\iiint \rho (\mathbf {r} )\delta (\mathbf {R} -\mathbf {r} )d\mathbf {r} =-\rho (\mathbf {R} )/\epsilon _{0},$

which proves that the Coulomb potential indeed satisfies the Poisson equation. Of course, a function F(R) that satisfies the Laplace equation

$\nabla ^{2}F(\mathbf {R} )=0,$

could have been added to Φ(R) and the result would still satisfy the Poisson equation. However, physically the presence of F(R) would indicate the presence of charges outside the volume covered by the triple integral. We assume that there are no such charges and hence that F = 0. (This argument can be rephrased as a boundary condition on Φ, we will omit this.)

Reference

J. D. Jackson, Classical Electrodynamics, 2nd edition, John Wiley, New York (1975).