electrostatics - Why, when and where is Gauss's law applicable?

Why is it said that Gauss's Law is mainly applicable for symmetric surfaces/bodies? Why not for asymmetric surfaces?

I want a logical explanation! BTW my teacher said that Gauss's law is applicable for any surface/body but in the case where symmetry does not exist, the calculation becomes a bit tedious. I did not get what he meant by that. Can someone please help me get a clear cut concept about when and where Gauss's law can be applied? Please note I'm not asking for the rigorous proof of Gauss's law.

Answer

The answer to your question involves the fact that one does not usually know a priori the electric field $\textbf{E}$ (or, for that matter, its direction) of a charge distribution $\rho$.

Gauss's law, in integral form, relates the flux of the electric field through some closed surface $S$ to the charge enclosed within the volume bounded by $S$. Precisely, it is the statement that given an electric field $\textbf{E}(\textbf{r})$ defined over space, the flux integral over any closed surface $S$ will always yield

$$ \oint_S \textbf{E} \cdot d\textbf{a} = \frac{Q_\text{enc}}{\epsilon_0}.$$

Normally surface integrals over vector fields involve parametrizing the surface (i.e. describing a two-dimensional surface by two parameters $u,v$ related to the Euclidean coordinates $x,y,z$). Even then, one has to additionally compute the product

$$\textbf{E} \cdot d\textbf{a} = \textbf{E} \cdot \hat{n}da,$$

where $\hat{n}$ is the unit normal to the surface and can be calculated from the parametrization. This quantity can assume different values everywhere along the surface.

So far I've only talked about the difficulties in computing the flux integral of a vector field over a general surface. When using Gauss's law, we have the added problem of not knowing the electric field (this is the quantity we're trying to find!). We now are tasked with computing an integral over an undefined function! This is where symmetry comes in and saves the troubled physicist.

Essentially, symmetric charge distributions allow one to choose a convenient surface (which preserves the symmetry) to remove $\textbf{E}$ from the integral. For example, consider a uniformly charged spherical volume of radius $R$ (i.e. a ball). Due to symmetry, one can argue that the electric field generated from this distribution must be radially symmetric. If we take our surface $S$ to be a sphere of radius $r$, then we find that the normal to the sphere and the direction of the electric field coincide, so

$$\textbf{E} \cdot d\textbf{a} = |\textbf{E}|\oint_S da = 4 \pi r^2 |\textbf{E}|,$$

since we are now simply computing the surface area of a sphere. We can now simply divide to find the answer:

$$ |\textbf{E}| = \frac{1}{4 \pi \epsilon_0} \frac{Q_\text{enc}}{r^2}.$$

To summarize, Gauss's (integral) law relates the flux integral of the electric field to the charge contained within a surface. Because we do not know the electric field, Gauss's law is only useful when we can remove the electric field from within the integral, which happens when the charge distribution displays certain spatial symmetries (spherical, cylindrical, planar).