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When we model electric field as something that represents the "flow" of something that is conserved, we can prove that the flux due to a single point charge, through the following surface,

Flux through a radial surface

is effectively zero. And therefore prove (almost) that flux through all kinds of surfaces that do not include a charge is zero. But when we do have a charge, we perform a clever trick, we enclose the charge with another surface, in the following manner,

Flux due to point charge

and therefore prove that the flux through the required surface is equal to the flux through small surface enclosing the charge. Therefore,

$$ \int_S \mathbf E . \mathbf {ds} = \dfrac{q}{4\pi \epsilon r^2} (4\pi r^2) \\ \int_S \mathbf E . \mathbf {ds} = \dfrac{q}{ \epsilon} $$

This "trick" was possible, only because, we had a point charge and the field was radially outward. So my question is, how are we able to generalise that the flux through any surface is equal to this fraction $\dfrac{q}{\epsilon}$? What if we had two point charges instead of one, then we wouldn't get a convenient $4\pi r^2$ term. This again leads to a question; why in the first place, is there a $4\pi$ in Coulomb's law?

P.S: I'm relatively new to the subject, so I'm missing something obvious.And I couldn't choose a suitable title for the question:|

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    $\begingroup$ Perhaps you should look up Gaussian units. link $\endgroup$ – Feel My Black Hole Jul 25 at 13:07
  • $\begingroup$ @FeelMyBlackHole So basically the factor $4 \pi$ is added, so as to avoid them in the "main"(Maxwell's) equations. So the proportionality constant in gaussian units is 1? $\endgroup$ – Aravindh Vasu Jul 25 at 13:12
  • $\begingroup$ You will need to understand the Divergence theorem to see why this $4\pi$ will show up in the more general case. $\endgroup$ – Danny Jul 25 at 15:10
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    $\begingroup$ @FeelMyBlackHole: Gaussian units don't help explain why Gauss' law holds for every closed boundary, or the factor of $4\pi$, since this factor appears either in Coulomb's law or (in the case of Gaussian units) Gauss' law. $\endgroup$ – Puk Jul 25 at 17:36
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    $\begingroup$ The fuss about units, Gaussian or otherwise, is a red herring. Ignore those comments. $\endgroup$ – garyp Jul 25 at 17:42
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Why does Gauss' law (in integral form) hold for any closed surface? One way of looking at it is that it just does: it's one of Maxwell's equations, and there is no "proving" it. In other words, there is experimental evidence for it. Many people find it more natural to express it in the differential form ($\vec{\nabla}·\vec{E} = \rho/\epsilon_0$), which is equivalent to the integral form (through the divergence theorem) but doesn't refer to a specific surface. Gauss' law can be motivated by Coulomb's law, or even derived from it in electrostatics, but Gauss' law is more fundamental in the sense that it holds in electrodynamics while Coulomb's law does not.

Here is a somewhat lazy derivation of Gauss' law from Coulomb's law using the divergence theorem that also illustrates how $4\pi$ comes about. In general it falls out of the math when dealing with situations of spherical symmetry.

From Coulomb's law, $$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\iiint\limits_\infty dV' \rho(\vec{r}') \frac{\vec{r} - \vec{r}'}{\lVert \vec{r} - \vec{r}' \rVert^3}.$$ Consider a region $V$ bounded by the closed surface $\partial V$. Then,

$$ \iint\limits_{\partial V}{\vec{E}(\vec{r})} · d\vec{S} =\frac{1}{4\pi\epsilon_0} \iint\limits_{\partial V}{ \iiint\limits_\infty dV' \rho(\vec{r}') \frac{(\vec{r} - \vec{r}')·d\vec{S}}{\lVert \vec{r} - \vec{r}' \rVert^3} }.$$

Changing the order of the integrals (which I won't attempt to justify), $$ \iint\limits_{\partial V}{\vec{E}(\vec{r})} · d\vec{S} =\frac{1}{4\pi\epsilon_0} \iiint\limits_{\infty}{ dV' \rho(\vec{r}') \iint\limits_{\partial V} { \frac{(\vec{r} - \vec{r}')·d\vec{S}}{\lVert \vec{r} - \vec{r}' \rVert^3} } }$$

Now consider the integral $$ I(\vec{r}') = \iint\limits_{\partial V} { \frac{(\vec{r} - \vec{r}')·d\vec{S}}{\lVert \vec{r} - \vec{r}' \rVert^3}} .$$ Suppose that $\vec{r}'$ is not enclosed by $\partial V$, so that the denominator is always non-zero and the integrand is always well defined. By the divergence theorem, $$ I(\vec{r}') = \iiint\limits_V {dV \vec{\nabla} · \frac{\vec{r} - \vec{r}'}{\lVert \vec{r} - \vec{r}' \rVert^3}} . $$ It's a straightforward calculation to show that the divergence in the integrand vanishes, therefore $I = 0$.

If on the other hand $\partial V$ encloses $\vec{r}'$, let $S_\delta$ be a sphere of radius $\delta$ centered about $\vec{r}'$ and contained in $\partial V$. Let $V - S_\delta$ be the subset of $V$ outside $S_\delta$, $\partial(V - S_\delta)$ the closed boundary of this region and $\partial S_\delta$ the spherical boundary of $S_\delta$. We can write $$ I(\vec{r}') = \iint\limits_{\partial(V - S_\delta)} { \frac{(\vec{r} - \vec{r}')·d\vec{S}}{\lVert \vec{r} - \vec{r}' \rVert^3}} + \iint\limits_{\partial S_\delta} { \frac{(\vec{r} - \vec{r}')·d\vec{S}}{\lVert \vec{r} - \vec{r}' \rVert^3}}. $$ The first integral is zero for the same reason as before: the divergence of the integrand vanishes everywhere. The second integral can now be evaluated easily using spherical symmetry to give

$$ I(\vec{r}') = \frac{\delta\times4\pi\delta^2}{\delta^3} = 4\pi.$$ This is where the factor of $4\pi$ comes from. Thus,

$$ \iint\limits_{\partial V}{\vec{E}(\vec{r})} · d\vec{S} =\frac{1}{4\pi\epsilon_0} \iiint\limits_{\infty}{ dV' \rho(\vec{r}') I(\vec{r}) } = \frac{1}{\epsilon_0} \iiint\limits_{V}{ dV' \rho(\vec{r}') } = \frac{Q}{\epsilon_0} $$ where the second equality uses the fact the $I(\vec{r}')$ is non-zero if and only if $V$ contains $\vec{r}'$.

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  • $\begingroup$ Can you please add a picture for $\partial (V-S_{\delta})$ and all that stuff. But ultimately it boils down to "math"? Is there a geometrical intuition behind the curtains? $\endgroup$ – Aravindh Vasu Jul 26 at 0:10
  • $\begingroup$ I'll describe in words what $\partial(V-S_\delta)$ looks like. Picture a simply connected region $V$, and the sphere $S_\delta$ enclosing $\vec{r}'$. Remove this sphere from $V$. What's left is the region $V$ with a "hole" in it. It now has an outer and an inner surface, $\partial(V-S_\delta)$ is the combination of these surfaces. The normal vectors on these surfaces point away from the bounded region, meaning it points outward on the outer surface and into the hole on the inner surface. $\endgroup$ – Puk Jul 26 at 0:35
  • $\begingroup$ As for geometric intuition, both in the derivation in my answer and when you apply Gauss' law to a spherically symmetric charge distribution (perhaps to derive Coulomb's law from Gauss' law, which is much easier), $4\pi$ appears because of spherical symmetry. This is because $4\pi$ is the solid angle subtended by a sphere. You can think of this factor as being a consequence of that fact that a point charge generates an E-field that looks the same whichever direction you look, i.e. a field that is spherically symmetric about the charge. $\endgroup$ – Puk Jul 26 at 0:37
  • $\begingroup$ So whenever we consider a number of charges, we consider such "holes" And conclude that the net flux out is equal to the flux out of all holes and finally add em up? $\endgroup$ – Aravindh Vasu Jul 26 at 0:49
  • $\begingroup$ Essentially yes. Basically if you have a number of point charges, the derivation says that the surface integral in Gauss' theorem is equal to the sum of surface integrals on spheres ("holes") enclosing each point charge. The contribution to the integral on each sphere comes only from the charge enclosed by the sphere. The result is Gauss' theorem. $\endgroup$ – Puk Jul 26 at 0:59
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Put a point charge inside a surface of any shape (or size). Consider a thin cone which extends out from the charge. Where the cone intersects the surface, the surface area is represented by a vector pointing outward. The associated flux is defined as the dot product of the E field and the area vector. We want the component of the field which is crossing the area, but that can also be interpreted as using the component of the area vector which is parallel to the field. This component of the area increases with the square of the distance as you move out along the cone, and the field decreases with the square of the distance. The flux out through the cone is a constant. Combine all the cones coming out from the charge and you get a constant total flux, which is independent of the size or shape of the surrounding surface. Choose a surface of constant radius, and the flux becomes E(4pi R^2). Each additional charge contributes its own flux.

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