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Could someone tell me please examples of physical situations (if there are) where there might appear a potential of the form

$$V(r)=V_{0}\ln(r/\xi),$$

being $r>0$ the radial coordinate in spherical 3D, $V_{0}$ is a constant and $\xi>0$ is a constant scale?

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  • $\begingroup$ This post (v1) seems like a list question. $\endgroup$ – Qmechanic Apr 12 '17 at 16:17
  • $\begingroup$ dear @Qmechanic which post? $\endgroup$ – user115376 Apr 12 '17 at 16:20
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    $\begingroup$ It looks like the potential inside a uniformly charged sphere with density prop. to $1/r^2$, so it diverges at the origin. $\endgroup$ – user126422 Apr 12 '17 at 16:57
  • $\begingroup$ $\log(r)$-type potential are typical of cylindrical, not spherical, symmetry. $\endgroup$ – ZeroTheHero Apr 12 '17 at 21:48
  • $\begingroup$ what about a mass with density 1/r^2, then the field is M/r^2=rho Vol/r^2=r^3/r^4=1/r, so the potential is ln(r) $\endgroup$ – user126422 Apr 13 '17 at 4:37
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Use Gauss' law in reverse. Imagine some material with permittivity $\epsilon$ and some spherically symmetric volume charge density $\rho(r)$. Take a Gaussian sphere of radius $R$ so that \begin{align} \oint_S \vec E\cdot d\vec S&= \vert \vec E\vert \times 4\pi R^2=\frac{1}{\epsilon}\oint_V\rho(r) dV\, ,\\ &=\frac{4\pi}{\epsilon} \int_0^R\,dr r^2\rho(r)\, . \end{align} As you want the field to go $\sim 1/R$ for the potential to be in $\log$, choosing $\rho(R)=\rho_0/r^2$ gives $$ \vert \vec E\vert r^2=\frac{\rho_0}{\epsilon} r $$ which implies $$ \vec E=\frac{\rho_0}{\epsilon r}\hat r $$ which in turn implies that, inside this material, the potential will go like $\log(r)$.

Note that, as pointed out by @ArmandoEstebanQuito, this would be a charge distribution singular at the origin. Also note that, if you wish to remove that singularity by having a hollow material so that $r=0$ is excluded, then $$ \int_0^{R} dr\,r^2\rho(r)\to \int_{R_0}^R dr\,r^2\rho(r) = \int_{R_0}^R dr = R-R_0 $$ so that now $$ \vert \vec E\vert = \frac{\rho_0}{\epsilon r^2}(r-R_0) $$ will contain a logarithmic part but also a $1/r$ part.

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  • $\begingroup$ @ArmandoEstebanQuito Oh dear! I just realized I repeated your comment, which I did not see because some comments were hidden for length of list. $\endgroup$ – ZeroTheHero Apr 13 '17 at 16:05

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