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10.6 Fields by Gauss' Law

10.6.1 Gauss' law

The flea of subsection 10.3.2 had a long and illustrious scientific career, and we're now going to pick up her story where we left off. This flea, whose name is Gauss9, has derived the equation \(E_\perp=2\pi k\sigma\) for the electric field very close to a charged surface with charge density \(\sigma\). Next we will describe two improvements she is going to make to that equation.

First, she realizes that the equation is not as useful as it could be, because it only gives the part of the field due to the surface. If other charges are nearby, then their fields will add to this field as vectors, and the equation will not be true unless we carefully subtract out the field from the other charges. This is especially problematic for her because the planet on which she lives, known for obscure reasons as planet Flatcat, is itself electrically charged, and so are all the fleas --- the only thing that keeps them from floating off into outer space is that they are negatively charged, while Flatcat carries a positive charge, so they are electrically attracted to it. When Gauss found the original version of her equation, she wanted to demonstrate it to her skeptical colleagues in the laboratory, using electric field meters and charged pieces of metal foil. Even if she set up the measurements by remote control, so that her the charge on her own body would be too far away to have any effect, they would be disrupted by the ambient field of planet Flatcat. Finally, however, she realized that she could improve her equation by rewriting it as follows:

\[\begin{equation*} E_{outward,\ on\ side\ 1}+E_{outward,\ on\ side\ 2} = 4\pi k\sigma . \end{equation*}\]

The tricky thing here is that “outward” means a different thing, depending on which side of the foil we're on. On the left side, “outward” means to the left, while on the right side, “outward” is right. A positively charged piece of metal foil has a field that points leftward on the left side, and rightward on its right side, so the two contributions of \(2\pi k\sigma\) are both positive, and we get \(4\pi k\sigma\). On the other hand, suppose there is a field created by other charges, not by the charged foil, that happens to point to the right. On the right side, this externally created field is in the same direction as the foil's field, but on the left side, the it reduces the strength of the leftward field created by the foil. The increase in one term of the equation balances the decrease in the other term. This new version of the equation is thus exactly correct regardless of what externally generated fields are present!

Her next innovation starts by multiplying the equation on both sides by the area, \(A\), of one side of the foil:

\[\begin{align*} \left(E_{outward,\ on\ side\ 1}+E_{outward,\ on\ side\ 2}\right)A &= 4\pi k\sigma A \\ \text{or} \\ E_{outward,\ on\ side\ 1}A+E_{outward,\ on\ side\ 2}A &= 4\pi kq , \\ \end{align*}\]

where \(q\) is the charge of the foil. The reason for this modification is that she can now make the whole thing more attractive by defining a new vector, the area vector A.


a / The area vector is defined to be perpendicular to the surface, in the outward direction. Its magnitude tells how much the area is.

 

As shown in figure a, she defines an area vector for side 1 which has magnitude \(A\) and points outward from side 1, and an area vector for side 2 which has the same magnitude and points outward from that side, which is in the opposite direction. The dot product of two vectors, \(\mathbf{u}\cdot\mathbf{v}\), can be interpreted as \(u_{parallel\ to\ v}|\mathbf{v}|\), and she can therefore rewrite her equation as

 

\[\begin{equation*} \mathbf{E}_1\cdot\mathbf{A}_1+\mathbf{E}_2\cdot\mathbf{A}_2 = 4\pi k q . \end{equation*}\]

The quantity on the left side of this equation is called the flux through the surface, written \(\Phi\).

b / Gauss contemplates a map of the known world.

 

Gauss now writes a grant proposal to her favorite funding agency, the BSGS (Blood-Suckers' Geological Survey), and it is quickly approved. Her audacious plan is to send out exploring teams to chart the electric fields of the whole planet of Flatcat, and thereby determine the total electric charge of the planet. The fleas' world is commonly assumed to be a flat disk, and its size is known to be finite, since the sun passes behind it at sunset and comes back around on the other side at dawn. The most daring part of the plan is that it requires surveying not just the known side of the planet but the uncharted Far Side as well. No flea has ever actually gone around the edge and returned to tell the tale, but Gauss assures them that they won't fall off --- their negatively charged bodies will be attracted to the disk no matter which side they are on.

Of course it is possible that the electric charge of planet Flatcat is not perfectly uniform, but that isn't a problem. As discussed in subsection 10.3.2, as long as one is very close to the surface, the field only depends on the local charge density. In fact, a side-benefit of Gauss's program of exploration is that any such local irregularities will be mapped out. But what the newspapers find exciting is the idea that once all the teams get back from their voyages and tabulate their data, the total charge of the planet will have been determined for the first time. Each surveying team is assigned to visit a certain list of republics, duchies, city-states, and so on. They are to record each territory's electric field vector, as well as its area. Because the electric field may be nonuniform, the final equation for determining the planet's electric charge will have many terms, not just one for each side of the planet:

\[\begin{equation*} \Phi = \sum \mathbf{E}_j\cdot\mathbf{A}_j = 4\pi k q_{total} \end{equation*}\]

 

Gauss herself leads one of the expeditions, which heads due east, toward the distant Tail Kingdom, known only from fables and the occasional account from a caravan of traders. A strange thing happens, however. Gauss embarks from her college town in the wetlands of the Tongue Republic, travels straight east, passes right through the Tail Kingdom, and one day finds herself right back at home, all without ever seeing the edge of the world! What can have happened? All at once she realizes that the world isn't flat.

c / Each part of the surface has its own area vector. Note the differences in lengths of the vectors, corresponding to the unequal areas.

 

Now what? The surveying teams all return, the data are tabulated, and the result for the total charge of Flatcat is \((1/4\pi k)\sum \mathbf{E}_j\cdot\mathbf{A}_j=37\ \text{nC}\) (units of nanocoulombs). But the equation was derived under the assumption that Flatcat was a disk. If Flatcat is really round, then the result may be completely wrong. Gauss and two of her grad students go to their favorite bar, and decide to keep on ordering Bloody Marys until they either solve their problems or forget them. One student suggests that perhaps Flatcat really is a disk, but the edges are rounded. Maybe the surveying teams really did flip over the edge at some point, but just didn't realize it. Under this assumption, the original equation will be approximately valid, and 37 nC really is the total charge of Flatcat.

d / An area vector can be defined for a sufficiently small part of a curved surface.

 

A second student, named Newton, suggests that they take seriously the possibility that Flatcat is a sphere. In this scenario, their planet's surface is really curved, but the surveying teams just didn't notice the curvature, since they were close to the surface, and the surface was so big compared to them. They divided up the surface into a patchwork, and each patch was fairly small compared to the whole planet, so each patch was nearly flat. Since the patch is nearly flat, it makes sense to define an area vector that is perpendicular to it. In general, this is how we define the direction of an area vector, as shown in figure d. This only works if the areas are small. For instance, there would be no way to define an area vector for an entire sphere, since “outward” is in more than one direction.

If Flatcat is a sphere, then the inside of the sphere must be vast, and there is no way of knowing exactly how the charge is arranged below the surface. However, the survey teams all found that the electric field was approximately perpendicular to the surface everywhere, and that its strength didn't change very much from one location to another. The simplest explanation is that the charge is all concentrated in one small lump at the center of the sphere. They have no way of knowing if this is really the case, but it's a hypothesis that allows them to see how much their 37 nC result would change if they assumed a different geometry. Making this assumption, Newton performs the following simple computation on a napkin. The field at the surface is related to the charge at the center by

 \[|\mathbf{E}| = \frac{kq_{total}}{r^2} ,\]

where \(r\) is the radius of Flatcat. The flux is then

\[ \Phi = \sum \mathbf{E}_j\cdot\mathbf{A}_j , \]

and since the \(\mathbf{E}_j\) and \(\mathbf{A}_j\) vectors are parallel, the dot product equals \(|\mathbf{E}_j||\mathbf{A}_j|\), so

\[\Phi = \sum \frac{kq_{total}}{r^2}|\mathbf{A}_j| .\]

But the field strength is always the same, so we can take it outside the sum, giving

\[\begin{align*} \Phi &= \frac{kq_{total}}{r^2} \sum |\mathbf{A}_j| \ &= \frac{kq_{total}}{r^2} A_{total} \ &= \frac{kq_{total}}{r^2} 4\pi r^2 \ &= 4\pi kq_{total} . \end{align*}\]

 

Not only have all the factors of \(r\) canceled out, but the result is the same as for a disk!

Everyone is pleasantly surprised by this apparent mathematical coincidence, but is it anything more than that? For instance, what if the charge wasn't concentrated at the center, but instead was evenly distributed throughout Flatcat's interior volume? Newton, however, is familiar with a result called the shell theorem (page 102), which states that the field of a uniformly charged sphere is the same as if all the charge had been concentrated at its center.10 We now have three different assumptions about the shape of Flatcat and the arrangement of the charges inside it, and all three lead to exactly the same mathematical result, \(\Phi = 4\pi kq_{total}\). This is starting to look like more than a coincidence. In fact, there is a general mathematical theorem, called Gauss' theorem, which states the following:

For any region of space, the flux through the surface equals \(4\pi kq_{in}\), where \(q_{in}\) is the total charge in that region.

Don't memorize the factor of \(4\pi\) in front --- you can rederive it any time you need to, by considering a spherical surface centered on a point charge.

Note that although region and its surface had a definite physical existence in our story --- they are the planet Flatcat and the surface of planet Flatcat --- Gauss' law is true for any region and surface we choose, and in general, the Gaussian surface has no direct physical significance. It's simply a computational tool.

Rather than proving Gauss' theorem and then presenting some examples and applications, it turns out to be easier to show some examples that demonstrate its salient properties. Having understood these properties, the proof becomes quite simple.

 

self-check:

Suppose we have a negative point charge, whose field points inward, and we pick a Gaussian surface which is a sphere centered on that charge. How does Gauss' theorem apply here?

(answer in the back of the PDF version of the book)

 

10.6.2 Additivity of flux

e / 1. The flux due to two charges equals the sum of the fluxes from each one. 2. When two regions are joined together, the flux through the new region equals the sum of the fluxes through the two parts.

 

Figure e shows two two different ways in which flux is additive. Figure e/1, additivity by charge, shows that we can break down a charge distribution into two or more parts, and the flux equals the sum of the fluxes due to the individual charges. This follows directly from the fact that the flux is defined in terms of a dot product, \(\mathbf{E}\cdot\mathbf{A}\), and the dot product has the additive property \((\mathbf{a}+\mathbf{b})\cdot\mathbf{c}=\mathbf{a}\cdot\mathbf{c}+\mathbf{b}\cdot\mathbf{c}\).

To understand additivity of flux by region, e/2, we have to consider the parts of the two surfaces that were eliminated when they were joined together, like knocking out a wall to make two small apartments into one big one. Although the two regions shared this wall before it was removed, the area vectors were opposite: the direction that is outward from one region is inward with respect to the other. Thus if the field on the wall contributes positive flux to one region, it contributes an equal amount of negative flux to the other region, and we can therefore eliminate the wall to join the two regions, without changing the total flux.

10.6.3 Zero flux from outside charges

A third important property of Gauss' theorem is that it only refers to the charge inside the region we choose to discuss. In other words, it asserts that any charge outside the region contributes zero to the flux. This makes at least some sense, because a charge outside the region will have field vectors pointing into the surface on one side, and out of the surface on the other. Certainly there should be at least partial cancellation between the negative (inward) flux on one side and the positive (outward) flux on the other. But why should this cancellation be exact?

To see the reason for this perfect cancellation, we can imagine space as being built out of tiny cubes, and we can think of any charge distribution as being composed of point charges. The additivity-by-charge property tells us that any charge distribution can be handled by considering its point charges individually, and the additivity-by-region property tells us that if we have a single point charge outside a big region, we can break the region down into tiny cubes. If we can prove that the flux through such a tiny cube really does cancel exactly, then the same must be true for any region, which we could build out of such cubes, and any charge distribution, which we can build out of point charges.

For simplicity, we will carry out this calculation only in the special case shown in figure f, where the charge lies along one axis of the cube. Let the sides of the cube have length \(2b\), so that the area of each side is \((2b)^2=4b^2\). The cube extends a distance \(b\) above, below, in front of, and behind the horizontal \(x\) axis. There is a distance \(d-b\) from the charge to the left side, and \(d+b\) to the right side.

f / The flux through a tiny cube due to a point charge.

 

There will be one negative flux, through the left side, and five positive ones. Of these positive ones, the one through the right side is very nearly the same in magnitude as the negative flux through the left side, but just a little less because the field is weaker on the right, due to the greater distance from the charge. The fluxes through the other four sides are very small, since the field is nearly perpendicular to their area vectors, and the dot product \(\mathbf{E}_j\cdot\mathbf{A}_j\) is zero if the two vectors are perpendicular. In the limit where \(b\) is very small, we can approximate the flux by evaluating the field at the center of each of the cube's six sides, giving

\[\begin{align*} \Phi &= \Phi_{left}+4\Phi_{side}+\Phi_{right} \\ &= |\mathbf{E}_{left}||\mathbf{A}_{left}|\cos 180° +4|\mathbf{E}_{side}||\mathbf{A}_{side}|\cos \theta_{side} \\ & +|\mathbf{E}_{right}||\mathbf{A}_{right}|\cos 0° , \text{and a little trig gives} \\ \text{$\cos\theta_{side}\approx b/d$, so} \\ \Phi &= -|\mathbf{E}_{left}||\mathbf{A}_{left}| +4|\mathbf{E}_{side}||\mathbf{A}_{side}|\frac{b}{d} +|\mathbf{E}_{right}||\mathbf{A}_{right}|\\ &= \left(4b^2\right)\left(-|\mathbf{E}_{left}| +4|\mathbf{E}_{side}|\frac{b}{d} +|\mathbf{E}_{right}|\right)\\ &= \left(4b^2\right)\left(-\frac{kq}{(d-b)^2} +4\frac{kq}{d^2}\frac{b}{d} +\frac{kq}{(d+b)^2}\right)\\ &= \left(\frac{4kqb^2}{d^2}\right)\left(-\frac{1}{(1-b/d)^2} +\frac{4b}{d} +\frac{1}{(1+b/d)^2}\right) .\\ \text{Using the approximation} \\ (1+\epsilon)^{-2}\approx 1-2\epsilon \\ \text{for small $\epsilon$, this becomes} \\ \Phi &= \left(\frac{4kqb^2}{d^2}\right)\left(-1-\frac{2b}{d} +\frac{4b}{d} +1-\frac{2b}{d}\right) \\ &= 0 . \end{align*}\]

Thus in the limit of a very small cube, \(b\ll d\), we have proved that the flux due to this exterior charge is zero. The proof can be extended to the case where the charge is not along any axis of the cube,11 and based on additivity we then have a proof that the flux due to an outside charge is always zero.

 

Example 35: No charge on the interior of a conductor

I asserted on p. 523 that for a perfect conductor in equilibrium, excess charge is found only at the surface, never in the interior. This can be proved using Gauss's theorem. Suppose that a charge \(q\) existed at some point in the interior, and it was in stable equilibrium. For concreteness, let's say \(q\) is positive. If its equilibrium is to be stable, then we need an electric field everywhere around it that points inward like a pincushion, so that if the charge were to be perturbed slightly, the field would bring it back to its equilibrium position. Since Newton's third law forbids objects from making forces on themselves, this field would have to be the field contributed by all the other charges, not by \(q\) itself. But this is impossible, because this kind of inward-pointing pincushion pattern would have a nonzero (negative) flux through the pincushion, but Gauss's theorem says we can't have flux from outside charges.

Discussion Questions

g / Discussion question A-D.

One question that might naturally occur to you about Gauss's law is what happens for charge that is exactly on the surface --- should it be counted toward the enclosed charge, or not? If charges can be perfect, infinitesimal points, then this could be a physically meaningful question. Suppose we approach this question by way of a limit: start with charge \(q\) spread out over a sphere of finite size, and then make the size of the sphere approach zero. The figure shows a uniformly charged sphere that's exactly half-way in and half-way out of the cubical Gaussian surface. What is the flux through the cube, compared to what it would be if the charge was entirely enclosed? (There are at least three ways to find this flux: by direct integration, by Gauss's law, or by the additivity of flux by region.)

◊ The dipole is completely enclosed in the cube. What does Gauss's law say about the flux through the cube? If you imagine the dipole's field pattern, can you verify that this makes sense?

◊ The wire passes in through one side of the cube and out through the other. If the current through the wire is increasing, then the wire will act like an inductor, and there will be a voltage difference between its ends. (The inductance will be relatively small, since the wire isn't coiled up, and the \(\Delta V\) will therefore also be fairly small, but still not zero.) The \(\Delta V\) implies the existence of electric fields, and yet Gauss's law says the flux must be zero, since there is no charge inside the cube. Why isn't Gauss's law violated?

◊ The charge has been loitering near the edge of the cube, but is then suddenly hit with a mallet, causing it to fly off toward the left side of the cube. We haven't yet discussed in detail how disturbances in the electric and magnetic fields ripple outward through space, but it turns out that they do so at the speed of light. (In fact, that's what light is: ripples in the electric and magnetic fields.) Because the charge is closer to the left side of the cube, the change in the electric field occurs there before the information reaches the right side. This would seem certain to lead to a violation of Gauss's law. How can the ideas explored in discussion question C show the resolution to this paradox?

10.6.4 Proof of Gauss' theorem

With the computational machinery we've developed, it is now simple to prove Gauss' theorem. Based on additivity by charge, it suffices to prove the law for a point charge. We have already proved Gauss' law for a point charge in the case where the point charge is outside the region. If we can prove it for the inside case, then we're all done.

If the charge is inside, we reason as follows. First, we forget about the actual Gaussian surface of interest, and instead construct a spherical one, centered on the charge. For the case of a sphere, we've already seen the proof written on a napkin by the flea named Newton (page 619). Now wherever the actual surface sticks out beyond the sphere, we glue appropriately shaped pieces onto the sphere.

h / Completing the proof of Gauss' theorem.

 

In the example shown in figure h, we have to add two Mickey Mouse ears. Since these added pieces do not contain the point charge, the flux through them is zero, and additivity of flux by region therefore tells us that the total flux is not changed when we make this alteration. Likewise, we need to chisel out any regions where the sphere sticks out beyond the actual surface. Again, there is no change in flux, since the region being altered doesn't contain the point charge. This proves that the flux through the Gaussian surface of interest is the same as the flux through the sphere, and since we've already proved that that flux equals \(4\pi kq_{in}\), our proof of Gauss' theorem is complete.

 

Discussion Questions

◊ A critical part of the proof of Gauss' theorem was the proof that a tiny cube has zero flux through it due to an external charge. Discuss qualitatively why this proof would fail if Coulomb's law was a \(1/r\) or \(1/r^3\) law.

10.6.5 Gauss' law as a fundamental law of physics

Note that the proof of Gauss' theorem depended on the computation on the napkin discussed on page 10.6.1. The crucial point in this computation was that the electric field of a point charge falls off like \(1/r^2\), and since the area of a sphere is proportional to \(r^2\), the result is independent of \(r\). The \(1/r^2\) variation of the field also came into play on page 622 in the proof that the flux due to an outside charge is zero. In other words, if we discover some other force of nature which is proportional to \(1/r^3\) or \(r\), then Gauss' theorem will not apply to that force. Gauss' theorem is not true for nuclear forces, which fall off exponentially with distance. However, this is the only assumption we had to make about the nature of the field. Since gravity, for instance, also has fields that fall off as \(1/r^2\), Gauss' theorem is equally valid for gravity --- we just have to replace mass with charge, change the Coulomb constant \(k\) to the gravitational constant \(G\), and insert a minus sign because the gravitational fields around a (positive) mass point inward.

Gauss' theorem can only be proved if we assume a \(1/r^2\) field, and the converse is also true: any field that satisfies Gauss' theorem must be a \(1/r^2\) field. Thus although we previously thought of Coulomb's law as the fundamental law of nature describing electric forces, it is equally valid to think of Gauss' theorem as the basic law of nature for electricity. From this point of view, Gauss' theorem is not a mathematical fact but an experimentally testable statement about nature, so we'll refer to it as Gauss' law, just as we speak of Coulomb's law or Newton's law of gravity.

If Gauss' law is equivalent to Coulomb's law, why not just use Coulomb's law? First, there are some cases where calculating a field is easy with Gauss' law, and hard with Coulomb's law. More importantly, Gauss' law and Coulomb's law are only mathematically equivalent under the assumption that all our charges are standing still, and all our fields are constant over time, i.e., in the study of electrostatics, as opposed to electrodynamics. As we broaden our scope to study generators, inductors, transformers, and radio antennas, we will encounter cases where Gauss' law is valid, but Coulomb's law is not.

10.6.6 Applications

Often we encounter situations where we have a static charge distribution, and we wish to determine the field. Although superposition is a generic strategy for solving this type of problem, if the charge distribution is symmetric in some way, then Gauss' law is often a far easier way to carry out the computation.

Field of a long line of charge

Consider the field of an infinitely long line of charge, holding a uniform charge per unit length \(\lambda\). Computing this field by brute-force superposition was fairly laborious (examples 10 on page 574 and 13 on page 580). With Gauss' law it becomes a very simple calculation.

The problem has two types of symmetry. The line of charge, and therefore the resulting field pattern, look the same if we rotate them about the line. The second symmetry occurs because the line is infinite: if we slide the line along its own length, nothing changes. This sliding symmetry, known as a translation symmetry, tells us that the field must point directly away from the line at any given point.

i / Applying Gauss' law to an infinite line of charge.

 

Based on these symmetries, we choose the Gaussian surface shown in figure i. If we want to know the field at a distance \(R\) from the line, then we choose this surface to have a radius \(R\), as shown in the figure. The length, \(L\), of the surface is irrelevant.

The field is parallel to the surface on the end caps, and therefore perpendicular to the end caps' area vectors, so there is no contribution to the flux. On the long, thin strips that make up the rest of the surface, the field is perpendicular to the surface, and therefore parallel to the area vector of each strip, so that the dot product occurring in the definition of the flux is \(\mathbf{E}_j\cdot\mathbf{A}_j=|\mathbf{E}_j||\mathbf{A}_j||\cos\ 0°=|\mathbf{E}_j||\mathbf{A}_j|\). Gauss' law gives

\[\begin{align*} 4\pi k q_{in} &= \sum \mathbf{E}_j\cdot\mathbf{A}_j \\ 4\pi k \lambda L &= \sum |\mathbf{E}_j||\mathbf{A}_j| .\\ \text{The magnitude of the field is the same on every strip,}\\ \text{so we can take it outside the sum.} \\ 4\pi k \lambda L &= |\mathbf{E}| \sum |\mathbf{A}_j| \\ \text{In the limit where the strips are infinitely narrow,}\\ \text{the surface becomes a cylinder, with (area)=(circumference)(length)=$2\pi RL$.} \\ 4\pi k \lambda L &= |\mathbf{E}| \times 2\pi RL \\ |\mathbf{E}| &= \frac{2k\lambda}{R} \\ \end{align*}\]

 

Field near a surface charge

As claimed earlier, the result \(E=2\pi k\sigma\) for the field near a charged surface is a special case of Gauss' law. We choose a Gaussian surface of the shape shown in figure j, known as a Gaussian pillbox. The exact shape of the flat end caps is unimportant.

j / Applying Gauss' law to an infinite charged surface.

 

The symmetry of the charge distribution tells us that the field points directly away from the surface, and is equally strong on both sides of the surface. This means that the end caps contribute equally to the flux, and the curved sides have zero flux through them. If the area of each end cap is \(A\), then

\[\begin{align*} 4\pi k q_{in} &= \mathbf{E}_1\cdot\mathbf{A}_1+\mathbf{E}_2\cdot\mathbf{A}_2 , \\ \text{where the subscripts 1 and 2 refer to the two end caps.} \\ \text{We have $\mathbf{A}_2=-\mathbf{A}_1$, so}\\ 4\pi k q_{in} &= \mathbf{E}_1\cdot\mathbf{A}_1-\mathbf{E}_2\cdot\mathbf{A}_1 \\ 4\pi k q_{in} &= \left(\mathbf{E}_1-\mathbf{E}_2\right)\cdot\mathbf{A}_1 , \\ \text{and by symmetry the magnitudes of the two fields are equal, so}\\ 2|\mathbf{E}|A &= 4 \pi k \sigma A\\ |\mathbf{E}| &= 2\pi k\sigma \end{align*}\]

 

The symmetry between the two sides could be broken by the existence of other charges nearby, whose fields would add onto the field of the surface itself. Even then, Gauss's law still guarantees

\[\begin{align*} 4\pi k q_{in} &= \left(\mathbf{E}_1-\mathbf{E}_2\right)\cdot\mathbf{A}_1 , \quad \text{or} \quad |\mathbf{E}_{\perp,1}-\mathbf{E}_{\perp,2}| = 4\pi k \sigma , \end{align*}\]

where the subscript \(\perp\) indicates the component of the field parallel to the surface (i.e., parallel to the area vectors). In other words, the electric field changes discontinuously when we pass through a charged surface; the discontinuity occurs in the component of the field perpendicular to the surface, and the amount of discontinuous change is \(4\pi k \sigma\). This is a completely general statement that is true near any charged surface, regardless of the existence of other charges nearby.

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