
#1
Apr1912, 06:27 AM

P: 128

I need a mathematical proof that should indicate the following: The direction of the electric field must be radial, for a spherical charge distribution to remain invariant after applying a rotation matrix to its field.
Analogously how can we prove that the electric field of a infinite cylinder is perpedicular to its surface? Moreover, consider our augmentation at the boundary value problem of electrostatics. Dirichlet's condition was once more taken as granted, of course there not due to some symmetry but nevertheless the principle remains the same! In other words, why is the electric field of the symmetric objects the way it is? Is it an experimental result or just the result of a superposition, which could be deducted? After all, all the great consequences of the laws of electrostatics are based on these initial "symmetry" assumptions! E.g. we could not calculate the electric field of a sphere with Gauss's law without "guessing" its direction! I need maths in order to unravel this symmetry mystery... 



#2
Apr1912, 03:27 PM

P: 4,664





#3
Apr1912, 03:41 PM

P: 312





#4
Apr1912, 04:04 PM

P: 4,664

Proof of symmetries in electrostatics 



#5
Apr2012, 12:51 PM

P: 128

@Bob S I don't mean to become ungrateful, but what I need to know is what I'm asking for (bold text). I'm already aware of the physical explanation.
I'm pretty sure, that a theorem of the following form can be found in the bibliography: D[itex]\vec{E}[/itex](D[itex]\vec{r}[/itex]) = [itex]\vec{E}[/itex]([itex]\vec{r}[/itex]) [itex]\Rightarrow[/itex] [itex]\vec{E}[/itex]([itex]\vec{r}[/itex]) = E([itex]\vec{r}[/itex])[itex]\vec{e_{r}}[/itex] That's what I'm looking for! 



#6
Apr2012, 09:10 PM

P: 4,664

I think first that you have to state that the surface of the sphere is conducting, and that the charge is static (no surface currents). Then using J = σE , you can show that if there are no surface currents, E_{θ}=E_{φ}=0. Thus there can be only a radial electric field E_{r}.
Then you can use the 3 components of [itex] \left(\nabla\times \overrightarrow{E} \right) =0 \space [/itex] to show that [itex] \frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \varphi}=0 [/itex]. Thus E_{r} is the same everywhere on the surface of the sphere: Curl E in spherical polar coordinates is [tex] \left(\nabla\times \overrightarrow{E} \right)_\theta=\frac{1}{r\sin\theta}\frac{\partial E_r}{\partial \varphi} \frac{1}{r}\frac{\partial\left(r E_\varphi \right)}{\partial r}=0 [/tex] [tex] \left(\nabla \times \overrightarrow{E}\right)_\varphi=\frac{1}{r}\left(\frac{\partial\left( r E_\theta \right)}{\partial r} \frac{\partial E_r}{\partial\theta } \right) =0[/tex] [tex] \left(\nabla \times \overrightarrow{E}\right)_r = \frac{1}{r\sin \theta}\left(\frac{\partial\left(\sin\theta E_\varphi \right) }{\partial \theta}\frac{\partial E_\theta}{\partial \varphi} \right)=0 [/tex] 



#7
Apr2212, 03:50 PM

P: 128

Well, you've mathematically elaborated your argument, namely the fact that the condition for an electrostatic equilibrium leads to the determination of the direction of the electric field.
That is appreciated. BUT, you should be aware that there is also another proof of this, involing only symmetries rotation matrixes and reflections on planes... I would have to compromise myself, that I'll never see that beautiful proof again 



#8
Apr2212, 07:39 PM

P: 312

If the charge is rotationally symmetric, so is the field. Therefore, the field cannot have tangential components. (For any tangential component at any point, a rotation along the axis connecting that point and the origin will change that component.) For cylinder, instead of considering rotational symmetry, consider reflection symmetries about the xy plane and any plane containing z axis, similar conclusion can be made about the tangential components.
BTW, the OP didn't even mention there's a conductor. 



#9
Apr2212, 10:33 PM

P: 404

[itex]\nabla . E(r,\phi,\theta)=\rho(r)/\epsilon[/itex]
According to my calculation , if we define vector filed F as : [itex]F(r,\phi,\theta)=\frac{\partial E(r,\phi,\theta)}{\partial \phi }[/itex] then we have [itex]\nabla . F=0[/itex] and [itex]\nabla × F=0[/itex] which mean [itex]\frac{\partial E}{\partial \phi }=0[/itex] I think you can prove the same for [itex]\frac{\partial E}{\partial \theta }[/itex]. 


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