# Solved Gauss' Law for $\vec{E}$ without boundary conditions?

Why can I solve for the electric field of a point charge Q at the origin without boundary conditions?

$\nabla\cdot\vec{E}=\rho/\varepsilon_0 = \delta(\vec{r})/\varepsilon_0$ is a 1st order differential equation, so it should need one boundary condition.

Through the divergence theorem you get Gauss' Law:

\begin{equation} \oint_S {E \cdot dA = \frac{Q}{{\varepsilon _0 }}} \end{equation}

By symmetry, the electric field is constant over any spherical surface surrounding the point charge Q. This simplifies the integral to:

\begin{equation} |E|(4\pi r^2)=Q/\varepsilon_0 \end{equation} or \begin{equation} \vec{E}=\frac{1}{4\pi \varepsilon_0}\frac{Q}{r^2}\hat{r} \end{equation} (the direction, $\hat{r}$, can again be implied by the symmetry of the problem).

When (if ever) did I give boundary conditions?

• I answered this question because I had some trouble with it at first and wanted to share. For those who down voted, is my answer incorrect in any way? Jan 8 '16 at 19:13
• Maybe some people object to answering your own question, but for the record, it is totally fine. (As long as the question and answer are acceptable on their own merits, of course.) Jan 9 '16 at 1:57
• Got it, thanks. As long as I haven't misinformed myself or others. Jan 9 '16 at 15:46

The point at which boundary conditions are specified is subtle. You can find this point by adding a constant vector to $$\vec{E}$$ in each step until it varies the equation: it should not change any equation until the boundary is specified. Since the flux of a vector field is unaffected by an added vector constant, the boundary condition was first specified when $$\vec{E}$$ was assumed to be spherically symmetric everywhere, which caused it to fall out of the integral.
It carries extra information that Maxwell's Equations alone do not: it assumes that the system is rotationally invariant. The boundary condition was that $$\vec{E}(r, \theta, \phi) = \vec{E}(r, \theta+\theta_0, \phi+\phi_0)$$ for any $$\theta_0, \phi_0$$, if putting it into an equation makes it seem more legitimate.