The gauss law tell you that the flux over an arbitrary closed surface around your body is proportional to the total charge: $$\int_{\partial V} \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0} $$ but this is an information about a peculiar integral of the field, and it is not easy to deduce the field at a given point without additional information.

Mar 1, 2015 · So Gauss' law is just an expression, in a different form, of the Coulomb law of forces between two charges. In fact, working back from Gauss' law, you can derive Coulomb's law. The two are quite equivalent so long as we keep in mind the rule that the forces between charges are radial.

Aug 24, 2016 · So, I know that the gauss law states that the Flux of the electric field through a closed surface is Q/ε , but does the gauss theorem works also for non inverse square law Fields? I think not because in order to not have a Flux depending on distance but a constant one we need that r^2 of the surface has to cancel with something that is r^2 too ...

Oct 1, 2014 · In the case of an electrostatic condition, Gauss's law can always be applied. Now, it doesn't mean that you can solve it analytically every time, because that is a different issue. We use Gauss's law in a highly symmetric situation because we can solve for the field quicker and easier than by the brute force method of Coulomb's law.

Aug 28, 2006 · Gauss's law in differential form states that the divergence of the electric flux density is equal to the volume charge density at that point. We can apply the integral form of Gauss's law and find the net outward flux through a surface enclosing a charge distribution, and then take the limit of the outward flux as the volume approaches zero.

Gauss' Law of electrostatics is an amazing law. It is extremely useful (as far as problems framed for it are concerned :D. I do not have a real world-problem solving experience of using Gauss' Law). However, I don't really understand why it works. What my …

holds, and take the divergence on both sides of the equation. This leads to the differential form of Gauss's law: $$\nabla\cdot E=\frac{\rho}{\epsilon_0},$$ where $\rho$ is the charge density. In order to reproduce its integral form, just integrate both sides of the equation. Detailed calculations can be found on the wikipedia page on Gauss's ...

Aug 18, 2011 · Gauss's law applies in all cases, but is only useful in cases with sufficient symmetry. As clem said, a finite disk has the wrong kind of symmetry. They probably meant a uniform surface charge density, usually symbolized as σ.

Edit: Also, another problem I noticed was that even if we remove the negative plate from the capacitor and then apply Gauss's Law in the same manner, the field still comes out to be $\sigma/\epsilon_0$ which is clearly wrong since the negative plate contributes to the field. So, maybe the problem is in the application of Gauss's Law.

Jan 11, 2016 · Coulomb's law is only true if the charges are stationary, there are no changing magnetic fields, etc. But Gauss's law is true under all circumstances. So Gauss's law is more than just a consequence of Coulomb's law. $\endgroup$ –