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I could, for example calculate the electric field near a charged rod of infinite length using the classic definition of the electric field, and integrating the: $$ \overrightarrow{dE} = \frac{dq}{4 \pi \varepsilon_0r^2}\overrightarrow{a_r}$$ Right? So, why in some cases that it's possible to also use this method, do we choose Gauss' law. And moreover, from which criterion do we judge to solve with either way? I am aware of the requirements to use Gauss' law but I stil haven't understood what are the patterns I should notice that would direct me to use Gauss' law to solve a problem.

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Your average physicist is lazy. Constructively lazy, but lazy nonetheless. Constructively lazy people take the easiest way that will work... – dmckee Aug 10 '12 at 19:36
Me too! I just still haven't figure out how exactly to recognise the easiest way, the patterns I should notice that would direct me to Gauss' law. See? I'm that lazy! :) – Dimitris Tzortzis Aug 10 '12 at 19:38
"Constructively lazy"... I like that (although it bears remembering that it is entirely different from actual laziness). – David Z Aug 10 '12 at 19:45
up vote 4 down vote accepted

Gauss's law is always true (that is, numerically), but it's not always useful for calculating electric fields. It's only useful for calculating a charge distribution's electric field when certain symmetries (e.g. cylindrical, spherical, or planar) are present that allow the surface integral to be done very simply. It is not useful for calculating electric fields when these symmetries are not present in a problem.

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That's helpful. Thanks! This could be an answer to my question. – Dimitris Tzortzis Aug 10 '12 at 19:41

In order to use Gauss Law, the electric field must be constant on the surface of integration (often a sphere). If so, E can be taken out of the surface integral.

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