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Poisson's equation
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==== 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|''Ο''(''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|β<sup>2</sup>''Ο''}}. Note that for {{mvar|r}} much greater than {{mvar|Ο}}, <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=405β415 |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|''Ο''(''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''Ο''}} the relative error is smaller than one part in a thousand.<ref name="Oldham"/>
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