Q:

Electric field of spherical charge distribution

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?

A:

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. 

