Gauss' Law: Study with Video Lessons, Practice Problems & Examples

Gauss's Law establishes a direct relationship between electric flux and enclosed charge, stating that the net flux through a closed surface is proportional to the charge within it. The equation is Φnet = Qε₀. This principle simplifies calculations of electric fields and flux, allowing for straightforward problem-solving in electrostatics, including determining flux from charge and vice versa, as well as deriving electric field equations for point charges using symmetric Gaussian surfaces.

Gauss' Law

Hey guys. So in this video, I want to cover a concept called Gauss's Law, which is a super important topic in electricity, that you need to know now in some textbooks. It's even an entire chapter and it can be kind of confusing because it ties together a lot of concepts and ideas about charge and flux. What I want to show you by the end of this video is that Gauss's Law is really just a very straightforward relationship between flux and charge.

And I'm gonna show you the three main types of problems that you'll need to know when solving problems. So let's go ahead and get started here. I'm just gonna jump straight in. So this guy, a long time ago, did a lot of calculations between charges and fluxes, and what he came up with was this law. And basically what it says is that the net flux that goes through a closed surface, which is really important—the closed part—depends only on the charge that is enclosed within that surface.

So basically, here's what I mean, right? Imagine I have this little box here and in the center of this box, I have a charge. And I'm just gonna pretend that it's a positive charge. Now, this positive charge emits electric field lines that go to the surface of this box. And instead of using, you know, some equations like $e\cdot a\cdot \cos (\theta )$ or whatever to calculate the flux, basically what Gauss's law says is that the total amount of flux going through all of the surfaces, so I'm gonna call this ${\varphi }_{\text{net}}$ here, is directly proportional to how much charge is in the box. And the equation for this is pretty straightforward. It's ${\varphi }_{\text{net}}=\frac{Q_{\text{enclosed}}}{{ϵ}_0}$. So this $ϵ$ here is just 8.85 times ${10}^{-12}$, a constant that we've seen before.

What I want to do now is show you the three most common types of problems that you'll need when using Gauss's Law. We're just gonna jump straight into the first one here. The first one is where you're given some kind of charges and you're asked to calculate the flux. Let's take a look at this first example here. What is the net flux that goes through this surface $A$ over here. Alright, so we've got this little surface $A$. We've got these two charges, one is positive and one is negative. So, I want to kind of suffer through this a little bit with you. Show you why this is super useful. Now, pretend for a second we didn't actually have Gauss's Law equation. The only way we can calculate and define is by using an old equation, which is $E\cdot A\cdot \cos (\theta )$.

So what happens is these point charges will emit electric fields like this; they'll have electric field lines like this and we have to calculate at each point, what is the $a\cdot \cos (\theta )$. The problem is this $E$ depends on $r$, so it depends on the distance. And what happens is this surface over here is going to be at a higher distance than this piece right here, and it's gonna be higher than this one, or different than this one. And also what happens is that the normal vector may sometimes point in a different direction than your electric field lines. Notice how this $E$ and this $A$ don't point in the same direction. So the angle is gonna be constantly changing. What this means here is that this is basically impossible. We can't calculate $E\cdot A$ using $E\cdot A\cdot \cos (\theta )$.

But now that we have Gauss's law, we can have a much more straightforward relationship between flux and charge. So this is ${\varphi }_{\text{net}}$ and this equals $\frac{Q_{\text{enclosed}}}{{ϵ}_0}$. So, what this means here is that the net flux through the surface only is directly proportional to how much charge is inside this surface here. So, how much charge is in the surface? Well, I've got 5 Coulombs and a negative 3 Coulombs charge. So, if you group these together and combine them, the total amount of charge I have in the surface here is going to be 5 Coulombs plus negative 3 Coulombs, and you'll just get 2 Coulombs. So there's basically 2 Coulombs of charge inside of the surface. So we just pop that into this equation. So what Gauss's law says is that ${\varphi }_{\text{net}}$ is equal to $\frac{2}{8.85\times {10}^{-12}}$. So when you work this out, what you're going to get here is $2.26\times {10}^{11}$ Newton meters squared per Coulomb.

So this is the power of Gauss's Law. All you have to do in a surface here is just know how much charge there is. And then you can figure out how much flux the total amount of flux that goes to the surface. And you don't have to use this $E\cdot A\cdot \cos (\theta )$ equation anymore. Alright, so, let's move on to the second problem now, which is kind of reversed. In some cases, you'll be given the flux either through one or multiple surfaces and you'll be asked for the charge. So, for example, we've got the flux through four sides of a closed pyramid and we've got these numbers over here.

So, I'm just gonna draw this out really quickly. Imagine I have this little pyramid like this. And basically, right, so, I've got, you know, 1, 2, 3, and then 4 sides, the underside like this. It doesn't matter which one is labeled, which it really doesn't matter because remember that Gauss's Law is only concerned with the net flux, not the ones through individual surfaces. So here, what we want to do in this problem if we want to calculate, well, what's the charge that's enclosed? Now, now we actually have the net flux and we want to figure out $Q$. So all we have to do here, I'm actually just gonna go ahead and move this down over here is figure out, Well, what is the net flux? Well, if you were told that the flux through each one of the surfaces, then the net flux is just gonna be adding all of them up.

So I'm just gonna add $51$ plus $52$ plus $53$ plus $54$. That's the net flux. So this just means that your net flux here is equal to we've got $10$ plus $20$ plus $8$. So this is $8$ plus negative $15$. And you're just gonna get $23$ this is Newton meters squared per Coulomb. So what I'm gonna do here is I'm just gonna replace my ${\varphi }_{\text{net}}$ with $23$ then I have to multiply by this, my $ϵ$, which is $8.85\times {10}^{-12}$. And this is gonna be my $Q_{\text{enclosed}}$. And when you work this out, what you're gonna get is $2.04\times {10}^{-}$ Coulombs.

So this is kind of the opposite here, you're using the flux to work backward and figure out. Well, if I know all the fluxes and that means that there's this amount of charge that's inside of the pyramid. I don't know how it's all arranged, it could be on the surface, it could be in the center. It doesn't matter. All I know here is that this must be the total amount of charge that's enclosed within that surface. Alright, now let's move on to the last problem here, the last kind of problem, which is a little bit more tricky. So in some problems, you may be given some kind of charges or a charge and you might be asked to calculate the electric field.

Electric Field Within Spherical Conductor

What's up, guys? Let's get more practice with Gauss's Law by checking out this example together. So, we're supposed to use Gauss's Law to figure out what the electric field is inside of a spherical conductor with some charge, negative Q. Okay, so what I want to point out is that this conductor, remember that the property of a conductor means that any net charges will always move and distribute themselves along the surface of that conductor. So, if you have some conductor right here and it has some net charge, negative Q, basically all of these negative charges will always distribute themselves evenly among that surface because they want to maximize their distance away from each other.

So now, we also know what we're working with; the third type of problem regarding Gauss's Law, figuring out what the electric field is. So, we have to choose a Gaussian surface with symmetry. Now, if you're working with a spherical sort of like conductor or spherical geometry thing, the best thing you're gonna use is also a sphere. But we have to use this field. We have to figure what the field is inside of this conductor. So the Gaussian surface that I'm gonna choose right here is gonna be within the conductor. Okay, so, basically, how do we relate this Gauss's Law to the electric field? We have that the total flux, which is e a times the cosine of theta is equal to the charge enclosed, so I can either use big Q or little q, Q_enclosed over Epsilon_0.

Now, what happens here? I've chosen this Gaussian surface within this conductor, and it's an arbitrary distance. So I'm choosing it; it could be here or here, over here somewhere. And I have to figure out what the enclosed amount of charge is. Now we said that for a conductor, all of the charges, all the net charges will always distribute themselves evenly among the surface. So that means that the amount of charge that's enclosed within the surface here is zero. There are no charges here. There are no charges within this Gaussian surface that I've chosen. There is a charge if I've chosen this Gaussian surface outside, but that's not what we're asked, were asked to figure out what the inside electric field is, so that means that this whole entire term is just equal to zero because Q_enclosed is equal to zero.

Now what happens is that, what does that mean for the electric field? We know that the area is not equal to zero because the area of our Gaussian surface is just gonna be whatever we choose that surface to be. It could be its radius or something like that. So what that means is that the electric field has to be equal to zero, and that actually makes sense. Think about when we're talking about the electric fields inside of conductors. We said the electric field inside a conductor always has to be zero. So Gauss's Law actually helps to reinforce this idea that if there are no charges within this Gaussian surface, then that means the electric field has to be zero. And that is the power of Gauss's Law. Let me know if you guys have any questions with this.

Rank the flux through surfaces A, B and C in the figure below from greatest to smallest.

• A. Φ_C, Φ_B, Φ_A
• B. Φ_A, Φ_B, Φ_C
• C. Φ_A, Φ_C, Φ_B
• D. Φ_B, Φ_C, Φ_A

Answer: (D)

Electric Field due to Hollow Shell

Hey, guys, let's go ahead and check out this problem. This is actually a very common type of problem that you'll see in Gauss's law. And it's very important that you know that. So watch this video a couple times, just in case you didn't catch everything. But basically what we're going to do is evaluate what the electric field is in these particular three regions surrounding this conducting spherical shell. One of them is going to be inside of the shell, but outside the point charge, one of them part B, is going to be within this conducting spherical shell, and then part C is going to be outside of this radius, this outer radius B here. So if we're trying to figure out what the electric field is, then one thing we can use is charges and we can use Gauss's law. Gauss's law helps us figure what the electric field is. But first, what we have to do is we have to draw a Gaussian surface and that Gaussian surface for a point charge. Hopefully, you guys know by now is going to be a perfect sphere. We covered that in the previous video. But basically the reason for that is that the electric field lines for this point charge point radiantly outwards. And that's also where the normals of the surfaces also point. So basically, these electric field lines and the normals always point in the same direction, no matter where you are on the Gaussian surface. So that means that if we're trying to evaluate the flux, which is going to tell us what the electric field is due to the charge that's enclosed, that's going to be the electric field times the area of this Gaussian surface times the cosine of theta. But what happens is because these things always point in the same direction. That means the cosine of the angle is just one, because theta at anywhere along the Gaussian surface between the electric field and the area vector, since they always point in the same direction, that theta is just equal to zero. So cosine zero is just one. Now, Gauss's Law tells us that the total amount of flux through this Gaussian surface that I've constructed is equal to the charge enclosed, divided by epsilon not, and what we're supposed to find is what this electric field is. So the only thing we have to do is figure out what the area of our Gaussian surface is. So I'm just going to go ahead and say that the Gaussian surface has a radius of little r. It's just some arbitrary thing I picked. If I picked it out here and drawn it a little bit wider, then that would be r right. So it's whatever basically you pick, and the area of that surface is just four pi r squared, because that's the surface area of a sphere. Now all we have to do is just move that area over to the other side and we get that the electric field is equal to Well, the charge enclosed is just going to be this little Q right here. So it's little Q divided by epsilon not, but we have to divide it by this one over four pi r squared. So one over four pi r squared, and we can actually rewrite this as one over four pi epsilon not times Q over r squared or another way that you might see that is just kQ over r squared. All of these are valid representations of the electric field, which is exactly what you would expect for just a point charge, because that's the enclosed charge of this Gaussian surface. It's just a point charge. Okay, so this is the answer to part A in part B, we're now going to be looking at inside of the spherical conducting shell. So we have to construct another Gaussian surface, and the Gaussian surface is going to be a sphere just like it was before. So now we can actually use a shortcut, because if we're trying to figure out the electric field and we know that we're inside of a conductor, remember how we talked about the electric field inside of a conductor is always equal to zero. This is an important fact that you must remember if you ever trying to value the electric field and you're inside of a conductor, that's going to be equal to zero. So some of you might be now confused. So if this is inside a conductor so this is the answer, by the way. Now, some of you guys might be really confused because you say the Gaussian surface that we have just constructed, right? If you imagine this Gaussian surface, what is the charge that's enclosed? It's this plus Q. So how could the electric field possibly be zero? Well, it turns out what happens is that the enclosed charge is equal to zero. Because even though you have this positive point charge that exists in the center of this gaussian surface, what ends up happening is that the electric field gets set up inside of this conductor so that there is a basically a distribution of charges that equals negative Q that gets set up inside of the inside walls of the conductor. So what happens is you have a plus Q from the point charge, right, so this plus Q is the point charge that's at the center. But what happens is that point charge basically sets up an electric field inside of the conductor that cancels it out. So this negative Q here is on the inside wall of this conductor. So that means that if you take a look at this Gaussian surface right now, the total amount of charge that's enclosed within this Gaussian surface is the negative Q. That's distributed evenly on the walls of the conductor and the positive Q. That's at the center of this Gaussian surface. So what's the total amount of charge that's enclosed? It's just zero. So that's why the electric field is zero. This is a very, very important thing. Watch that explanation over again. So the positive Q sets up a field inside of the conductor that cancels out the electric fields to the electric field to zero. And because of that, there's a distribution of negative Q that gets set up on the inside walls. Now what happens is that due to charge conservation, the fact that you can't just generate a negative Q out of nowhere, that means that a positive Q has to get set up on the outside walls. So this is what happens. It basically just polarizes these charges so that they cancel out the electric field and it's equal to zero. Alright, so go ahead and watch those videos on the electric field inside a conductor just so you make sense of that. Okay, so basically, the answer is that inside of the electric inside of the conductor, the electric field is equal to zero, as it always should be. Alright, in the last part, now, as we have to figure out what the E at part C is so r greater than B, right? So we have to set up our Gaussian surface. The Gaussian surface that we're going to use is going to be all the way out here, right? And it doesn't matter how far we make it. But what we do know is that this distance out here is just going to be r. And so this is the Gaussian surface that we've constructed. So that means that the phi, the total amount of electric flux is e a cosine of theta. Right? We know that theta is always just going to be equal to one. And all we have to do is just set this e a here is equal to the Q enclosed over epsilon not. So let's take a look. What is the amount of charge that's enclosed within our Gaussian surface? We have a plus Q that gets set up on the outer wall of the conductor. The inside of the wall gets a negative Q. And then we have this positive Q that's at the center of that sphere. So that means that this Q enclosed right here is just equal to Q minus Q plus Q. So let me go ahead and remove myself here. So I've got Q minus Q plus Q. So that means that these charges just go away and you end up with just a positive Q over here. So that means that the electric flux is just going to equal to, Let's see, the area of the spherical Gaussian surface that we've constructed is four pi r squared. Just it always was. And that means that the electric field, actually let's see, means you want to move this over. That means the electric field is equal to plus Q over epsilon not and then divided by one over the four pi r squared. So that means that that's just equal to kQ over r squared, just as it is for a point charge. So it turns out that this spherical conductor doesn't actually really do anything other than set up a charge or set up a distribution of charges so that it cancels out the electric field inside of this. But then anywhere else after that, it basically just starts to look like a point. Charge again. Right. So make sure you go and understand each one of these parts here. By the way, this is actually supposed to be part C over here. Let me know if any one of these processes or steps confused you. Just drop me a link in the comments and I'll answer your questions. All right, let me know if you have any questions. See you later.

Surface Charge Density

Alright, guys, welcome back. We're going to work out this example together. We're supposed to figure out what the surface charge density is on the inner and outer surfaces. Now, before you watch this, just make sure that you've seen the last video on how to figure out what the electric fields are at different points throughout this conducting surface. Because you need to know what happens with the charges that are set up on the inside walls, things like that. Just make sure you watch the last video and you're good to go. So we do know from the last video that due to Gauss's Law, there's an electric field inside of here. The electric field is zero within the conductor, and that exists again on the outside. And this is because whatever charge exists over here gets set up and basically distributed around the inner walls of the inside of this conductor.

Here, we know that if this is three Coulombs, then there has to be a negative three Coulombs that gets set up on the inner walls. And then on top of that, there has to be a positive three Coulombs that gets set up on the outer walls because of charge conservation, right? So I'm just going to use plus signs everywhere. So basically what we're supposed to do is figure out what the surface charge density is on these inner surfaces. Now, that surface charge density is given by the letter Sigma and that Sigma has an equation. It's the total amount of charge divided by the total amount of area.

I want to be very careful here because this Q is not the enclosed charge, like what we're using in Gaussian surfaces. So Q is not the Q enclosed. What we're basically doing is we're just figuring out, "Okay, if all of these charges here get distributed on the surface of this tiny little thin shell right here on the inside walls of this conductor, what is the total amount of charge divided by the total amount of area here?" So just don't make the mistake and say that this is the total amount of enclosed charge. That's different.

Σ = Q A , where Q is the total amount of charge and A is the area.

Sigma on the inside wall is going to be the total amount of charge that's on the inside wall, in other words, the negative three Coulombs, divided by the area of the inside wall. Now, the area of the inside wall here is just equal to 4 π R 2 inside squared, and we actually have what the surfaces are. We know this is three centimeters, and this is five centimeters on the outer wall. In other words, A _ inside is just 4 π , and now we have to convert that; we have to square that. So the area of the inside wall is going to be 1.13 × 10 - 2 in square meters.

So basically, I can go ahead and just plug that in here and figure out what the surface charge density is. So Sigma on the inside is going to be Q, which is negative three Coulombs, divided by 1.13 × 10 - 2 , and my sigma inside equals, let's see, negative 265 Coulombs per meter squared, right? So that's basically how you figure out what the surface charge density of the inside wall is.

To do the same thing on the outside wall is just going to be the exact same process. So, to figure out what Sigma out is, that's just going to be Q divided by A of the outside wall. So this A _ outside is just going to be 4 π , and then now we just use instead of using three centimeters, we're just going to use five centimeters. I'm going to plug that in. So I've got 0.5, and then we're going to square that. So that means that this outside area is just equal to 3.14 × 10 - 2 , and that's just going to go straight in this equation right here.

Okay, so that means that the outside surface charge density is just the let's see, we've got three Coulombs distribute on the outside divided by the area, which is 3.14 × 10 - 2 , and that is equal to 95 Coulombs per meter squared. Notice how this surface charge density ends up being positive, and this one ends up being negative because that actually has to do with the amount of charge that's divided by this area here. So those charges actually could be negative or positive. Alright, let me know if you guys have any questions with this, I'll see you in the next one.

A spherical, thin conducting shell of radius 8cm has a charge of –6C. If a 4C charge were placed at the center of the shell, what is the electric field 4 cm from the center? At 12 cm?

• A. E_4cm = 2.25 × 10⁹ outwards;E_12cm = 1.25 × 10⁸ inwards
• B. E_4cm = 2.25 × 10¹³ outwards;E_12cm = 1.25 × 10¹³ outwards
• C. E_4cm = 2.25 × 10¹³ outwards;E_12cm = 6.25 × 10¹² outwards
• D. E_4cm = 2.25 × 10¹³ outwards;E_12cm = 1.25 × 10¹² inwards

Answer: (D)

Here’s what students ask on this topic:

What is Gauss's Law and how is it used in electrostatics?

Gauss's Law states that the net electric flux through a closed surface is proportional to the charge enclosed within that surface. The mathematical expression is:

$\Phi$_net = Qε₀

where $\Phi$_net is the net flux, $Q$ is the enclosed charge, and $\epsilon$₀ is the permittivity of free space. This law simplifies the calculation of electric fields and flux, especially for symmetric charge distributions. By choosing an appropriate Gaussian surface, one can easily relate the enclosed charge to the electric field, making it a powerful tool in electrostatics.

How do you calculate electric flux using Gauss's Law?

To calculate electric flux using Gauss's Law, follow these steps:

1. Identify the closed surface (Gaussian surface) that encloses the charge.
2. Determine the total charge $Q$ enclosed within the surface.
3. Use the formula:

$\Phi$_net = Qε₀

4. Substitute the values of $Q$ and $\epsilon$₀ (8.85 × 10^-12 C²/N·m²) into the equation.
5. Calculate the net flux $\Phi$_net.

This method allows you to find the electric flux through a closed surface without needing to integrate the electric field over the surface.

What are the common types of problems solved using Gauss's Law?

There are three common types of problems solved using Gauss's Law:

1. Calculating Flux from Charge: Given a charge distribution, determine the net electric flux through a closed surface using $\Phi$_net = Qε₀.
2. Calculating Charge from Flux: Given the net electric flux through a closed surface, determine the enclosed charge using _netε₀.
3. Deriving Electric Field: Use Gauss's Law to derive the electric field for symmetric charge distributions (e.g., point charges, spherical, cylindrical, or planar symmetry) by choosing an appropriate Gaussian surface.

These problems highlight the versatility and power of Gauss's Law in simplifying complex electrostatic calculations.

How do you choose a Gaussian surface for solving problems using Gauss's Law?

Choosing a Gaussian surface involves selecting a surface that exploits the symmetry of the charge distribution to simplify calculations. The three main types of Gaussian surfaces are:

1. Spheres: Used for point charges or spherically symmetric charge distributions. The electric field is constant at a given radius.
2. Cylinders: Used for line charges or cylindrically symmetric charge distributions. The electric field is constant at a given radial distance from the axis.
3. Planes: Used for infinite plane charges or planar symmetry. The electric field is constant at a given distance from the plane.

By choosing a surface where the electric field is constant and the angle between the field and the surface normal is zero, you can simplify the flux calculation to .

What is the significance of the permittivity of free space (ε₀) in Gauss's Law?

The permittivity of free space, denoted as $\epsilon$₀, is a fundamental physical constant that characterizes the ability of the vacuum to permit electric field lines. Its value is approximately 8.85 × 10^-12 C²/N·m². In Gauss's Law, it appears in the denominator of the equation:

$\Phi$_net = Qε₀

This constant ensures that the relationship between electric flux and enclosed charge is correctly scaled. It plays a crucial role in determining the strength of the electric field generated by a given charge distribution in a vacuum.