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When solving the equation $$\boxed{ {1 \over r^2}{\partial \left( r^2 E_r \right) \over \partial r} + {1 \over r\sin\theta}{\partial \over \partial \theta} \left( E_\theta\sin\theta \right) + {1 \over r\sin\theta}{\partial E_\phi \over \partial \phi} = Const}$$ where $Const$ depends only on $r$ and $E_r$, $E_{\theta}$ and $E_{\phi}$ are three unknown functions and we assume the boundary conditions are spherical symmetric, what was the argument for $E_{\theta}$ and $E_{\phi}$ being zero?

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  • $\begingroup$ We can write this equation more simply as $\vec{\nabla} \cdot \vec{E} = f(\vec{r})$, and in general $E_\theta$ and $E_\phi$ will be non-zero. However, if the boundary conditions imposed also respect the spherical symmetry of the equation quoted, then the solution will also. That is, the solution will be invariant under rotation. $\endgroup$ – gj255 Sep 16 '14 at 15:11
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Only from that equation, you cannot deduce that the solution is spherically symmetric. Take, for example $E_\phi=r\sin(\theta)$ and $E_\theta=0$. Usually, you cannot solve this equation, or more precisely, there are many solutions. The point is that, in order to solve the electric field, you need to take into consideration the rest of the Maxwell equations. For example, if the problem is stationary, $\partial_t B=\partial_tE=0$, then $\nabla\times E=0$, and therefore $E=-\nabla\psi$. In consequence $-\Delta\psi=f(r)$. This last equation indeed implies that $\psi=\psi(r)$, and therefore $E$ is radial.

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  • $\begingroup$ I think you mean $\partial_t B = 0$? Also, I should point out, for those who haven't met the notation, that $\Delta \psi$ means exactly the same as $\nabla^2 \psi$. $\endgroup$ – gj255 Sep 17 '14 at 10:51
  • $\begingroup$ Both time derivatives. I assumed an electrostatic problem. $\endgroup$ – Enredanrestos Sep 17 '14 at 12:10
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The argument is "boundary conditions are spherically symmetrical".

If you have a non zero value for the tangential component, then you have to lose the symmetry.

Spherical symmetry means you can rotate in any direction and get the same answer. That is only true if the field is the same regardless of the value of $\theta$ or $\phi$. You should, specifically, be able to rotate about any point on the surface on a line through the center: now a tangential component will rotate as well. Only if that component is zero do you not lose the symmetry.

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