Editing Poisson's equation (section)

=== Electrostatics ===
{{main|Electrostatics}}
{{refimprove|data=September 2024|date=September 2024}}

Many problems in [[electrostatics]] are governed by the Poisson equation, which relates the [[electric potential]]
{{mvar|φ}} to the free charge density 
<math>\rho_f</math>, such as those found in [[Electrical conductor|conductors]]. 

The mathematical details of Poisson's equation, commonly expressed in [[SI units]] (as opposed to [[Gaussian units]]), describe how the [[charge density|distribution of free charges]] generates the electrostatic potential in a given [[region (mathematics)|region]].

Starting with [[Gauss's law]] for electricity (also one of [[Maxwell's equations]]) in differential form, one has
<math display="block">\mathbf{\nabla} \cdot \mathbf{D} = \rho_f,</math>
where <math>\mathbf{\nabla} \cdot</math> is the [[divergence|divergence operator]], '''D''' is the [[electric displacement field]], and ''ρ<sub>f</sub>'' is the free-[[charge density]] (describing charges brought from outside).

Assuming the medium is linear, isotropic, and homogeneous (see [[polarization density]]), we have the [[constitutive equation#Electromagnetism|constitutive equation]]
<math display="block">\mathbf{D} = \varepsilon \mathbf{E},</math>
where {{mvar|ε}} is the [[permittivity]] of the medium, and '''E''' is the [[electric field]].

Substituting this into Gauss's law and assuming that {{mvar|ε}} is spatially constant in the region of interest yields
<math display="block">\mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho_f}{\varepsilon}.</math>
In electrostatics, we assume that there is no magnetic field (the argument that follows also holds in the presence of a constant magnetic field).<ref>{{Cite book |last=Griffiths |first=D. J. |year=2017 |title=Introduction to Electrodynamics |edition=4th |publisher=Cambridge University Press |pages=77–78}}</ref>
Then, we have that
<math display="block">\nabla \times \mathbf{E} = 0,</math>
where {{math|∇×}} is the [[Curl (mathematics)|curl operator]]. This equation means that we can write the electric field as the gradient of a scalar function {{mvar|φ}} (called the [[electric potential]]), since the curl of any gradient is zero. Thus we can write
<math display="block">\mathbf{E} = -\nabla \varphi,</math>
where the minus sign is introduced so that {{mvar|φ}} is identified as the [[electric potential energy]] per unit charge.<ref>{{Cite book |last=Griffiths |first=D. J. |year=2017 |title=Introduction to Electrodynamics |edition=4th |publisher=Cambridge University Press |pages=83–84}}</ref>

The derivation of Poisson's equation under these circumstances is straightforward. Substituting the potential gradient for the electric field,
<math display="block">\nabla \cdot \mathbf{E} = \nabla \cdot (-\nabla \varphi) = -\nabla^2 \varphi = \frac{\rho_f}{\varepsilon},</math>
directly produces Poisson's equation for electrostatics, which is
<math display="block">\nabla^2 \varphi = -\frac{\rho_f}{\varepsilon}.</math>

Specifying the Poisson's equation for the potential requires knowing the charge density distribution. If the charge density is zero, then [[Laplace's equation]] results. If the charge density follows a [[Boltzmann distribution]], then the [[Poisson–Boltzmann equation]] results. The Poisson–Boltzmann equation plays a role in the development of the [[Debye–Hückel equation|Debye–Hückel theory of dilute electrolyte solutions]].

Using a Green's function, the potential at distance {{mvar|r}} from a central point charge {{mvar|Q}} (i.e., the [[fundamental solution]]) is
<math display="block">\varphi(r) = \frac {Q}{4 \pi \varepsilon r},</math>
which is [[Coulomb's law]] of electrostatics. (For historical reasons, and unlike gravity's model above, the <math>4 \pi</math> factor appears here and not in Gauss's law.)

The above discussion assumes that the magnetic field is not varying in time. The same Poisson equation arises even if it does vary in time, as long as the [[Coulomb gauge]] is used. In this more general class of cases, computing {{mvar|φ}} is no longer sufficient to calculate '''E''', since '''E''' also depends on the [[magnetic vector potential]] '''A''', which must be independently computed. See [[Mathematical descriptions of the electromagnetic field#Maxwell's equations in potential formulation|Maxwell's equation in potential formulation]] for more on {{mvar|φ}} and '''A''' in Maxwell's equations and how an appropriate Poisson's equation is obtained in this case.

==== Potential of a Gaussian charge density ====
If there is a static spherically symmetric [[Gaussian distribution|Gaussian]] charge density
<math display="block">\rho_f(r) = \frac{Q}{\sigma^3\sqrt{2\pi}^3}\,e^{-r^2/(2\sigma^2)},</math>
where {{mvar|Q}} is the total charge, then the solution {{math|''φ''(''r'')}} of Poisson's equation
<math display="block">\nabla^2 \varphi = -\frac{\rho_f}{\varepsilon}</math>
is given by
<math display="block">\varphi(r) = \frac{1}{4 \pi \varepsilon} \frac{Q}{r} \operatorname{erf}\left(\frac{r}{\sqrt{2}\sigma}\right),</math>
where {{math|erf(''x'')}} is the [[error function]].<ref>{{Cite journal |last1=Salem |first1=M. |last2=Aldabbagh |first2=O. |title=Numerical Solution to Poisson's Equation for Estimating Electrostatic Properties Resulting from an Axially Symmetric Gaussian Charge Density Distribution |journal=Mathematics |volume=12 |issue=13 |pages=1948 |year=2024 |doi=10.3390/math12131948 |doi-access=free }}</ref> This solution can be checked explicitly by evaluating {{math|∇<sup>2</sup>''φ''}}.

Note that for {{mvar|r}} much greater than {{mvar|σ}}, <math display="inline">\operatorname{erf}(r/\sqrt{2} \sigma)</math> approaches unity,<ref name="Oldham">{{Cite book |last1=Oldham |first1=K. B. |last2=Myland |first2=J. C. |last3=Spanier |first3=J. |title=An Atlas of Functions |chapter=The Error Function erf(x) and Its Complement erfc(x) |pages=405–415 |year=2008 |publisher=Springer |location=New York, NY |doi=10.1007/978-0-387-48807-3_41 |isbn=978-0-387-48806-6 |chapter-url=https://link.springer.com/chapter/10.1007/978-0-387-48807-3_41}}</ref> and the potential {{math|''φ''(''r'')}} approaches the [[electrical potential|point-charge]] potential,
<math display="block">\varphi \approx \frac{1}{4 \pi \varepsilon} \frac{Q}{r},</math>
as one would expect. Furthermore, the error function approaches 1 extremely quickly as its argument increases; in practice, for {{math|''r'' > 3''σ''}} the relative error is smaller than one part in a thousand.<ref name="Oldham"/>