# Gauss theorem and inverse square law [duplicate]

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?

## marked as duplicate by Aaron Stevens, John Rennie, ZeroTheHero, Qmechanic♦Dec 30 '18 at 6:46

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## 1 Answer

I'd like to draw a distinction:

Gauss's Theorem (Also called Divergence Theorem):

$$\iint_{\partial V} \mathbf{E} \cdot d\mathbf{A} = \iiint_V \nabla \cdot \mathbf{E}\ dV$$

This is a purely mathematical statement and holds for all differentiable vector fields $$\mathbf{E}$$.

Gauss's Law: $$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

Plugging this into Gauss's Theorem we have that

$$\iint_{\partial V} \mathbf{E} \cdot d\mathbf{A}= \frac{1}{\epsilon_0} \iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{enc}}{\epsilon_0}.$$

So to answer your question, Gauss's Theorem is always true. It must be. However, Gauss's Law didn't have to be true; it just so happens to be a law of physics in the universe we find ourselves in.

Having said that, Gauss's law will be true for any vector field $$\mathbf{F}$$ that satisfies the differential equation.

$$\nabla \cdot \mathbf{F} \propto \Lambda(\mathbf{r}),$$

where $$\Lambda$$ is just some scalar field that is well defined over the volume $$V$$. “Gauss's law” (i.e. in integral form; I put it in quotes because we are just plugging it into Gauss’s theorem, but this is how many use the term) for such a field would look like

$$\iint_{\partial V} \mathbf{F} \cdot d\mathbf{A}= \alpha \iiint_V \Lambda(\mathbf{r}) \ dV \equiv \alpha \tilde{Q}_{enc},$$

for some constant $$\alpha$$ and $$\tilde{Q}_{enc}$$ is just what we define to be how much "charge" is enclosed by our surface.

Conclusion: The fact that the $$\mathbf{E}$$ field falls like $$\frac{1}{r^2}$$ just makes the integral on the left hand side do-able at a fixed $$r$$ (actually the fact that it only depended on $$r=|\mathbf{r}|$$ is what made it do-able); it does not mean that anything that the corresponding field $$\mathbf{F}$$ falls like $$\frac{1}{r^2}$$. When you think about all the times you actually used Gauss's law to calculate the $$\mathbf{E}$$ field recall that you actually had to assume that it fell like $$\frac{1}{r^2}$$ in order to do the integral simply.