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An electrostatic field is characterized by the fact that it depends only by $r$, isn't it? If it is true, I don't understand why this expression, given in cylindrical coordinates,

$${\bf E(r)}=\frac{\alpha}{z^2}{\bf u_r}-2 \frac{\alpha r}{z^3}{\bf u_z} $$

rapresents an electrostatic field.

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  • $\begingroup$ The electrostatic fields due to a single point charge has spherical symmetry. That doesn't means that the field due to other charge distributions are required to obey the same symmetry. $\endgroup$
    – dmckee --- ex-moderator kitten
    Sep 15, 2013 at 20:28
  • $\begingroup$ @dmckee how can I be sure that an expression rapresents an electrostatic fiel? $\endgroup$
    – sunrise
    Sep 15, 2013 at 20:31
  • $\begingroup$ @PhysiXxx your expression is the electrostatic fiel given by a single point charge? $\endgroup$
    – sunrise
    Sep 15, 2013 at 20:58
  • $\begingroup$ @sunrise. I deleted my comment because it doesn't refer to general case. But the field of rest charge is also electrostatic, so your expression is also must be right for it. But it doesn't. In general, your expression ignores charge distribution. $\endgroup$
    – user8817
    Sep 15, 2013 at 21:01
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    $\begingroup$ @sunrise . Maybe, this expression refer to some geometrical figure with constant charge density. It is electrostatic field, because it isn't time-dependence. $\endgroup$
    – user8817
    Sep 15, 2013 at 21:08

1 Answer 1

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An electric field is said to be static if it does not change with time, i.e. the the charges that produced that field are stationary. This doesn't imply any constriction on its spatial dependence. In particular, no spherical symmetry is implicit in the definition of electrostatic field, and that field may not depend only on $r$, as your example shows. This is common when you consider examples of field produced by more than one point charge, e.g. an electric dipole: $$\mathbf{E}(\mathbf{r})={3\mathbf{p}\cdot\hat{\mathbf{r}}\over 4\pi\varepsilon_0 r^3}\hat{\mathbf{r}}-{\mathbf{p}\over 4\pi\varepsilon_0 r^3}$$ depends on $\mathbf{r}$ and $\mathbf{p}$.

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    $\begingroup$ many thanks! You are right, I haven't thought it! And if I would know which charge distribution gives this expression of E, how can I do? $\endgroup$
    – sunrise
    Sep 15, 2013 at 21:10
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    $\begingroup$ Well, you can use the differential form of the Gauss's law: $\boldsymbol{\nabla}\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}$ which implies (if I did it right): $\rho=\alpha{6r\over z^4}.$ $\endgroup$
    – AstoundingJB
    Sep 15, 2013 at 21:23
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    $\begingroup$ Uhm... I forgot a factor $\varepsilon_0$, the vacuum permittivity, at the numerator on the right..! I can't find the edit button..! $\endgroup$
    – AstoundingJB
    Sep 15, 2013 at 21:31
  • $\begingroup$ wow! :) Probably I have done a mistake because I don't obtain your result for $\rho$... $\endgroup$
    – sunrise
    Sep 15, 2013 at 21:37
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    $\begingroup$ Ok, let's check it! In cylindrical coordinate, the gradient operator is $\boldsymbol{\nabla}=\left({\partial\over\partial r},{1\over r}{\partial\over\partial \theta},{\partial\over\partial z}\right),$ where I'm using the (radial, azimuthal, vertical) coordinates $(r,\theta,z)$, right? $\endgroup$
    – AstoundingJB
    Sep 15, 2013 at 21:44

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