In this lesson, we'll derive Guass's law. We proved in the previous section that the electric flux through any surface \(S\) produced by any charge distribution is given by

$$\phi_E=∮\vec{E}·d\vec{A}.\tag{1}$$

The equation above is very general; but suppose that we considered the following special case: the electric flux through any closed surface produced by any point charge \(q_1\) which is somewhere inside of that closed surface. In Figure 1, I have drawn a green, blob-shaped, closed surface; I chose such a wierd looking surface to try to emphasize the fact that we could have chosen any closed surface: a sphere, cylinder, a weird blob shaped enclosure, and so on. The point charge \(q_1\) could be at any arbitrary location, as long as it is inside the closed surface \(S\). The question is: what is the electric flux through this surface? To answer this question, we just need to solve Equation (1) in the special case in which we're dealing with the electric flux through any arbitrary surface produced by jsut a single point charge. We showed, in an earlier lesson, that using Column's law and the definition of an electric field, that the electric field produced by a single point change (in this case \(q_1\)) is given by:

$$\vec{E}=k\frac{q}{r^2}\vec{r}.\tag{2}$$

Let's substitute this result into Equation (1) to get

$$\phi_E=∮k\frac{q}{r^2}\vec{r}·d\vec{A}.\tag{3}$$

So far so good. But there's just one problem: the integral in Equation (3) would be very hard to solve. Fortunately, there's a way to make life easier. Equation (3) applies to any closed surface. Let's choose a sphere centered around the point charge \(q_1\) as illustrated in Figure 2. Let me explain what's nice about a sphere. Every term \(\vec{E}·d\vec{A}\) in the infinite sum \(∮\vec{E}·d\vec{A}\) corresponds to the dot product between \(\vec{E}\) and \(d\vec{A}\) at every point on the sphere. Every electric field vector generated by \(q_1\) will be perpendicular to the surface and therefore parallel or anti-parallel to the vector \(d\vec{A}\). If \(q_1\) is positive, then every electric field vector \(\vec{E}\) will be parallel and pointing in the same direction as \(d\vec{A}\). Therefore, every term \(\vec{E}·d\vec{A}\) in the infinite sum will just simplify to (\vec{E}·d\vec{A}=EdA\). Thus, we have

$$\phi_E=k\frac{q}{r^2}∮dA.\tag{4}$$

The integral \(∮dA\) is the infinite sum of the areas of each surface element. This just gives the total area \(A\) of the sphere. We already know that the surface area of a sphere is \(4πr^2\); if we make this substitution into Equation (4), then we have

$$\phi_E=k\frac{q}{r^2}4πr^2=4πkq=\frac{q}{\epsilon_0}.\tag{5}$$

Equation (5) is a very special case of Guass's Law for a point charge. But since the true practical usefulness of Guass's Law is its generality and that it can be applied to any charge distribution (not just a point charge), let's generalize this proof to any charge distribution.

Suppose that we have a sphere enclosing an \(n\) number of point charges. The electric flux through the sphere is given by

$$\phi_E=∮(\vec{E_1}+...+\vec{E_n})·d\vec{A}=∮\vec{E_1}·d\vec{A}+...+∮\vec{E_1}·d\vec{A}.\tag{6}$$

Just as a reminder, we want to find the total electric flux through \(S\). The only way to do this is to add up the total flux through every surface element and to get the total flux through each element, we must add up the contributions due to all the other point charges. That's why the integrand is the way it is and involves the sum, \(\vec{E_1}+...+\vec{E_n}\), in which all the electric fields produced by each point charge get accounted for.

What's nice about this problem is that we already finished the hard part: namely, finding the electric flux due to a single point charge. This allows us to simple Equation (6) to

$$\phi_E=∮\vec{E}·d\vec{A}=\frac{q_1+...+q_n}{\epsilon_0}=\frac{q_{enc}}{\epsilon_0}\tag{7}$$

where \(q_{enc}\) is equal to \(\sum_{i=1}^nq_i\) and is the total charge distribution enclosed inside of the sphere. We can generalize this result (the result that \(\phi_E= \frac{q_{enc}}{\epsilon_0}\)) to any closed surface enclosing the charge distribution (not just a sphere) by being clever. No matter what closed surface you choose that encloses all of the point chages, the same number of electric field lines will pass through that surface as the sphere. Given the empirical facts that the total electric flux \(\phi_E\) through a surface is proportional to the number of field lines passing through the surface (which we can write as \(\phi_E∝N\)), it follows that the electric flux \(\phi_E\) must be the same through any surface enclosing the charge.

And just to make a finish remark and to complete this generalization, we proved this for any system of \(n\) charged particles. We could let the number \(n\) of these charges particles approach infinity in which case we'll be considering any continuous charge distribution. In this case, the proof would be exactly the same except for one difference which I'll write below:

$$\phi_E=∮\vec{E_1}·d\vec{A}+...+∮\vec{E_n}·d\vec{A}+...= \frac{q_1+...+q_n+...}{\epsilon_0}= \frac{q_{enc}}{\epsilon_0}$$

where the charge enclosed inside of the surface, \(q_{enc}=\lim_{n→ ∞}\sum_{i=1}^{n}q_i\), is made of infinitely made point charges and becomes a continuous charge distribution. What if there are charges outside of the surface? Would that effect this calculation? The answer is no. The electric field lines generated by all of the point charges outside of the surface will pass into the surface, but then pass out of the surface. The total electric flux is proportional to the difference of how many field lines pass out minus the number that come in. From this empirical fact, we can conclude that the electric flux through the arbitrary closed surface due to charges outside of the closed surface is zero and, therefore, we'll just end up with the same calculations that we did throughout this proof.

I'd like to summarize everything that we just did in a sentence: for any charge distribution, whether a system of a finite number of point charges or a continuous distribution of charges, if I draw any arbitrary closed surface (which can be of any shape, size, and location), then the only contributions to the electric flux through that closed surface will be due to only the charges \(q_{enc}\) enclosed within that surface and that electric flux is given by the amount \(\frac{q_{enc}}{\epsilon_0}\).

That proof was fun and all—albeit a little tedious. It's important to understand this proof since it gives us an intuition of how Guass's law is applied, when it is applicable and when it is not applicable. It is important to appreciate the word *any* to fully grasp the applicability of Guass's law: it applies to *any* closed surface and to *any *distribution of charge. In the next couple of lessors, we're going to begin to look at a lot of applications of Guass's law (starting off with simple problems) and we'll be dealing with many different kinds of closed surfaces and charge distributions. By starting off with the simple problems, we'll get a feel for how Guass's law is used in practice to solve problems. But eventually, we'll start to use Guass's law for very important applications. When we apply Guass's law in these applications, we'll see that Guass's law predicts some surprising relationships between electricity and conductors. This understanding which Guass's law gives us about how electricity and conductors are related to each other gives was very important for desinging and constructing numerous technologies which rely on electricity.