Good morning. Today we are going to review electric flux and Gauss’ law for AP Physics C: Electricity and Magnetism. ♪ Flipping Physics ♪ The term “flux” is defined as any effect that appears to pass or travel through a surface or substance, however, realize that effect does not need to actually move. Hence, the phrase “appears to” in the definition. Billy, please remind me of what “electric flux” for a constant electric field is. The symbol for flux is an uppercase Phi. When we add a capital E subscript, that makes it electric flux. Electric flux is the measure of the amount of electric field which passes through a defined area. The equation for electric flux of a uniform electric field through a two-dimensional area is uppercase Phi sub capital E equals the dot product of the uniform electric field and the area of the surface through which the electric field is passing where both electric field and area are vectors. That also equals electric field times area times the cosine of the angle between the direction of the electric field and the direction of the area. Area can have a direction? Yeah, the direction of an area is always normal to the plane of the area just like the direction of angular velocity is normal to the plane in which the object is rotating. Oh yeah. Right. This equation has the same form as the work equation for a constant force. Work equals the dot product of force and displacement or force times displacement times the cosine of the angle between the force doing the work and the displacement of the object. That means we should use the magnitudes of the electric field and area, and that the value of the cosine of the angle between the electric field and the area determines if the electric flux is positive or negative. Electric flux is a scalar. The units for electric flux are well, electric field is in newtons per coulomb and area is in square meters. So, the units for electric flux are newtons times square meters divided by coulombs. Thank you Billy. Now, usually when we determine electric flux it is through some sort of closed surface. So, let’s do an example and determine the net electric flux of a uniform, horizontal electric field through a right triangular box oriented as I have shown in the illustration. To determine the net electric flux, we need to define some dimensions of the right triangular box. Let’s label the short side of the right triangle, a, the long side, b, the hypotenuse, c, and the width of the right triangular box, w. In order to determine the net electric flux through the closed surface of the right triangular box, we need to determine the electric flux through all five sides of the right triangular box and then sum all those electric fluxes. The plural of flux is fluxes and not just flux? The plural of tux is tuxes. But the plural of influx is influx or influxes. The plural of fox is foxes. But the plural of ox is ox. Or oxen. But the plural of sock is just socks, not sockses. Sock ends in c k not x. Oh, right. Fun fact. When my daughter Geneve was learning to talk she deduced that the singular of box was bock. That makes sense. Sure. Bock. Yeah. So Now we need to identify all five sides of the right triangular bock. … (Nice. Yeah. Right.) Let’s identify the left, rectangular side as side 1, in red. The bottom as side 2, in yellow. The triangular side closest to us as side 3, in pink. The triangular side farthest from us as side 4, in blue. And the top side of the right triangular box, which is formed with the hypotenuse of the right triangle and the width of the box, as side 5 in black. (The colors kind of work.) (Eh, it works well enough. It’s not that complicated.) (Sure.) Good. Bobby, please determine the electric flux through area 1, the back of the right triangular box. Okay. Area 1, the back. Well, we use the equation for electric flux from a constant electric field which equals let’s not use the dot product because we do not have unit vectors electric field times area 1 times the cosine of angle 1. Area 1 is a rectangle, so its area equals length a times length w. And the area vector for area 1. Is that area vector to the left or to the right? I know the area is normal to the plane of the surface, however, that could be to the left or to the right. (Area vectors are always directed outward for closed surfaces.) Area vectors are always out of closed surfaces. That means the area vector for area 1 is to the left. The electric field is to the right and the angle between to the left and to the right is 180 degrees. The cosine of 180 degrees is negative one. So, the electric flux through area 1 equals negative electric field times side a times side w. (Thank you Bobby. Bo, how about the electric flux through area 2, the bottom?) Sure. Area 2, the bottom. We use the same equation Bobby just used, only with 2’s for subscripts instead of 1’s. Area 2 equals b times w. The direction of area 2 is well, you said the area vector is always out of the closed surface, that means area 2 is directed down. That makes the angle between area vector 2, which is down, and the electric field vector, which is to the right, 90 degrees. And the cosine of 90 degrees is zero. There is zero electric flux through the bottom of the right triangular box. That makes sense, if you look at the diagram, the electric field lines are parallel to the bottom surface of the right triangular box, so none of the electric field lines go through the bottom surface of the right triangular box, so the electric flux through the bottom surface of the right triangular box equals zero. (Bobby, the electric flux through side 3, the right triangular surface which is closest to us, please?) Sure. Side 3. The triangle closest to us. Using the same equation with subscripts of 3. The area is the area of a triangle or one-half base, which is b, times height, which is a, and then times the cosine of the angle between area 3 and the electric field. Area 3 is out of the right triangular box or out of the screen. The angle between out of the screen and to the right is again 90 degrees. So, yeah, the electric flux through area 3 is zero. No electric field lines pass through area 3. Actually, the same is true for area 4, the triangle which is farthest from us. No electric field lines pass through area 4, so the electric flux through area 4 is also zero. And the area vector for area 4 is into the screen, and the angle between into the screen and to the right is 90 degrees, so the math for the electric flux through area 4 is also gives us zero flux. Thank you Bobby. Now, the electric flux through area 5, the top surface of the right rectangular box requires one more illustration to understand. The area of area 5 is a rectangle, which has area of length c times length w. The angle between area vector 5 and the electric field takes a bit of geometry to understand. You can see angle 5 in the illustration. By definition, the direction of area vector 5 and the hypotenuse of the triangle form a right angle. That makes the angle between the dotted extended line for the electric field and the hypotenuse of the right triangle 90 degrees minus angle 5. Because the angle between the horizontal dotted line and the vertical side of the triangle is 90 degrees, that makes the top angle in the right triangle also equal to angle 5. (Oh, cool.) Therefore, the cosine of angle 5, which equals adjacent over hypotenuse, also equals length a over length c. Which we can substitute back into the equation for electric flux to replace cosine of angle 5. Length c cancels out, and the electric flux through side 5 equals the magnitude of the electric field, times length a, times length w. That means the net flux through the closed right triangular box equals the sum of all 5 of those fluxes. … (Again with the fluxes.) Which equals, when you add all of them together, zero. Uh? Okay. Excuse me. I need a sip of my tea. Sure you do. Nice. I don’t get it. You’ve never had tea before… Alright. Please notice that, when an electric field is going into a closed surface, the electric flux is negative, and, when an electric field is coming out of a closed surface, the electric flux is positive. It’s just useful to realize that. Okay, let’s do another example. Let’s determine the electric flux passing through a sphere which is concentric to and surrounds a positive point charge. Bo, please work on this one. Well, You just gave us the equation for electric flux. Electric flux equals the dot product of the electric field and the area, so let’s use that. We know the electric field is … We can’t use that equation. Why not? That is the equation for electric flux when the electric field is uniform. The electric field caused by a point charge is not uniform. The electric field decreases as you get farther from the point charge. Yeah. Uh… We need to identify infinitesimally small areas of the surface, which are dA, and the infinitesimally small fluxes dPhi created by the electric field passing through dA. Actually, rather than using “fluxes” for the plural of “flux”, I am going to say “electric flux values”. Of course you are. Yes. Take the integral of both sides. And we get an equation for electric flux which does not need the electric field to be constant. Electric flux equals the integral of the dot product of electric field and dA. Okay. Let’s use that equation. However, let’s use E dA cosine theta instead of the dot product. The angle between the electric field and area dA is, for every area dA, equal to zero. That is because the electric field is always outward and dA is always outward; they are both in the same direction. The cosine of zero is 1. Actually, the electric field has the same magnitude as it passes through every infinitesimally small area, dA, on the concentric, spherical surface through which we are finding the flux. It does? Yeah, Coulomb’s law gives the electrostatic force as the Coulomb constant times charge 1 times charge 2 all divided by the square of the distance between centers of charge all times unit vector r. The electric field equals the electrostatic force divided by the charge on the small, positive test charge. That means the electric field around a positive point charge equals the Coulomb constant times the test charge times the point charge all divided by the square of the distance between centers of charge and all of that divided by the test charge. The charge of the test charge cancels out, and the electric field around a point charge equals the Coulomb constant times the charge of the point charge, all divided by the square of the distance between the center of the point charge and the location of the electric field. In other words, the electric field which surrounds and is caused by a point charge is constant at a constant radius. The concentric sphere has a constant radius, so the electric field there is constant. So, we can take electric field out from under the integral. In other words, we could have just used the equation I started with. Maybe. But this is, mathematically, a more rigorous approach. Sure. Okay. The integral of dA is A; the area of the concentric spherical surface. The surface area of a sphere equals four pi r squared. We can substitute in the equation for the electric field which surrounds a point charge and the equation for the surface area of a sphere into the electric flux equation. The square of the distance from the center of the point charge to the electric field cancels out. And, that looks good to me. We also know the Coulomb constant equals the inverse of the quantity four pi times the permittivity of free space. four pi cancels out. And we get a more elegant answer. The electric flux which passes through a sphere which is concentric to a positive point charge equals the charge of the positive point charge divided by the permittivity of free space. Sure. A more elegant answer using a more mathematically rigorous approach. So, the electric flux is independent of the radius of the concentric spherical surface? Cool! That is weird. Yeah. Yep. And, what we have just shown here is a demonstration of Gauss’ law. Gauss’ law states that the electric flux through a closed surface equals the closed surface integral of the dot product of electric field and area dA which equals the charge enclosed in the Gaussian surface divided by the permittivity of free space. In the example we just did, capital Q is the charge of the point charge, and it is the charge enclosed inside an imaginary spherical surface and the imaginary spherical surface is called a Gaussian surface. That circle in the integral sign is confusing. It just means the integral is over a closed surface. It’s called a closed surface integral. Okay. I guess that helps. Thanks. Sure. More specifics about using Gauss’ law. A Gaussian surface is a three-dimensional closed surface. And, realize, while the Gaussian surfaces we usually work with are imaginary, the Gaussian surface could actually be a real, physical surface. Typically, we choose the shapes of our imaginary Gaussian surfaces such that the electric field generated by the enclosed charge is either perpendicular to or parallel to the sides of the Gaussian surface. This greatly simplifies the surface integral because all the angles are multiples of 90 degrees and the cosine of those angles have a value of -1, 0, or 1. You can see that we did this in our previous example. Because the Gaussian surface we chose was a sphere concentric with the point charge, the electric field and the area vector were always in the same direction giving an angle of zero degrees in the dot product. If we had chosen a cube as our imaginary Gaussian surface, the angle between the electric field and the area vector would have had all sorts of different values. And the electric field would not have been constant at the Gaussian surface because it was a cube. Correct Billy. And, as long as the amount of charge enclosed in a Gaussian surface is constant, the total electric flux through the Gaussian surface does not depend on the size of the Gaussian surface. Again, we showed this with the example we just did. And, Gauss’ law is the first of Maxwell’s equations which are a collection of equations which fully describe electromagnetism. We will get to the rest of Maxwell’s equations in due time. I will also point out that, if the net charge inside a closed Gaussian surface is zero, then the net electric flux through the Gaussian surface is also zero. Oh, that is why the net electric flux through the right triangular box was zero. That is why you drank your tea! Yeah. Thanks Bobby. Let’s do another example using Gauss’ law. Let’s determine the electric field which surrounds an infinitely large, thin plane of positive charges with uniform surface charge density, lowercase sigma. Basically, we have an infinitely large, infinitesimally thin, two-dimensional plane of positive charges which are evenly spread out through the infinite plane. It’s like a really, really large sheet of paper with a uniform, positive charge per unit area. And our job is to use Gauss’ law to determine the electric field which surrounds and is caused by the infinite plane of positive charges. First off, we know the electric field will be normal to and directed away from the infinite plane of positive charges. This is because the plane is infinitely large; therefore, every component of the electric field, dE, which is parallel to the plane of charges and is caused by infinitesimally small, charged pieces of the plane, dq, will cancel out leaving only electric field components of dE which are perpendicular to the plane and directed away from the plane of charges. Now that we have that figured out, Bobby, what shape shall we choose for our imaginary Gaussian surface? Well, for the shape of the imaginary Gaussian surface, we are trying to make sure the electric field is either parallel or perpendicular to all the surfaces of the Gaussian surface. The electric field is always horizontal, so the shape of the Gaussian surface needs to have sides which are vertical or horizontal. Could we use a cylinder that has a horizontal axis and has its center on the plane of charges? Bobby, we certainly can. So, our charges are represented by this piece of paper, and our Gaussian surface is represented by this can of soup. That is our visual representation. We spare no expense here at Flipping Physics. We can afford demonstrations just like this because of your Patreon donations. Sorry, I know that may have sounded a little sarcastic. I truly do appreciate everyone who donates monthly on Patreon and all donations through other avenues as well. Honestly, I would not be able to make these videos and keep them for free for everyone without your financial help. Thanks. Billy, please solve for the electric field in this case. Absolutely. Uh … The imaginary closed Gaussian cylindrical surface is made up of 3 surfaces. The 2 ends of the cylinder and the side of the cylinder which wraps all the way around the cylinder. That means the electric flux through the imaginary closed Gaussian cylindrical surface equals the electric flux through the side of the cylinder plus the electric flux through the left end of the cylinder plus the electric flux through the right end of the cylinder. Rather than using the dot product equation, let’s use the electric flux equation without the dot product. The angle for the side of the Gaussian cylindrical surface is between a horizontal electric field and the direction of any dA on the side which is always 90 degrees. That means, because the cosine of 90 degrees equals zero, there is zero electric flux through the side of the Gaussian cylinder which makes sense because the electric field lines do not actually pass through the side of the Gaussian cylinder. The angle between the horizontal electric field and the direction of dA for the left end of the Gaussian cylinder is actually the angle for both ends of the Gaussian cylinder are always equal to zero degrees because the area vectors for each of the ends of the Gaussian cylinder are outward and therefore in the same direction as the electric field. The cosine of zero degrees equals 1, and the electric field has the same value no matter which dA we are talking about, which means electric field can come out from under the integral. And the integral with respect to area is just area, however, there are two ends, the left and right ends, and they both have the same area, so that is 2 times the area of 1 end. And that all still equals the charge inside the Gaussian surface divided by the permittivity of free space. Now we need to deal with that right-hand side of the equation. We know surface charge density equals charge over area. When looking at just the charge enclosed by the imaginary Gaussian cylinder, we know the charge in that equation is the charge inside the imaginary Gaussian cylinder and the area of the charge inside the Gaussian cylinder is the area of one end of the cylinder. That means the charge inside the Gaussian cylinder equals the surface charge density times the area of one end. On the right-hand side of the equation we can substitute the surface charge density times the area of just one end. And … Everybody brought the area of one end of the Gaussian cylinder to the party! And we get that the magnitude of the electric field which surrounds and is caused by an infinitely large, thin plane of charges equals surface charge density divided by the quantity two times permittivity of free space. Very nice Billy. Let’s review a couple of things. Notice the electric field is uniform and independent of distance from the plane of charges. This is because the plane of charges is infinitely large. Also, I just want to review that the Gaussian surface we chose has Gaussian surface area vectors that are either parallel to or perpendicular to the electric field. Realize that means we could have chosen a rectangular box as our imaginary, closed Gaussian surface. That would have given the exact same electric field. And let’s take this example one step further. What if we add an equal magnitude, parallel, infinite plane of negative charges? Bo, what will happen then? Oh, and realize, the surface charge density in the equation now refers to the magnitude of the surface charge density, which is the same for both infinite planes of charges. Bo? Well, the electric field from the positively charged plane is uniform in magnitude everywhere. Before we add the electric field from the equal magnitude, negatively charged plane, let’s make the electric field from the positively charged plane blue. Actually, let’s remove the negatively charged plane for a moment so we can visualize what the electric field fully looks like from just the positively charged plane. Now let’s switch to just the negatively charged plane. The electric field from the negatively charged plane, let’s make it red, is also uniform everywhere, however, that electric field points towards the negatively charged plane because electric fields are defined using a small, positive test charge, and a positive charge would be attracted to the negatively charged plane. And, now let’s add back the positively charged plane and its blue electric field. Okay, outside the two parallel, oppositely charged planes, the electric fields are equal in magnitude, however, they are opposite in direction, therefore the electric field outside the two parallel planes of charge is zero. However, between the two parallel planes of charge, the two electric fields are in the same direction. Therefore, the electric field between the parallel planes of charge is just two times the electric field of one plane of charge. Substitute in surface charge density over two times the permittivity of free space and we get that the electric field that exists between two parallel planes of charge with equal magnitude, but opposite sign is equal to surface charge density over permittivity of free space. Well done Bo. Thanks. And realize, we have now begun our journey towards determining the capacitance of an ideal parallel plate capacitor. Cool. Yea? Yup. Also, notice that a positively charged particle moving through a constant electric field like this will experience an electrostatic force in the direction of the electric field. This force will be constant and equal to the charge on the particle times the magnitude of the electric field. In other words, the motion of a charged particle through a constant electric field will have similar characteristics to a mass moving through a constant gravitational field near the surface of a planet. Oh, that’s just like projectile motion! Exactly, this is very similar to projectile motion. Okay. A few loose ends to tidy up about electric flux and Gauss’ law. According to the AP Physics C: Electricity and Magnetism Guidelines, you are responsible for a quantitative approach of Gauss’ law to solve for electric fields only for charge distributions which are spherically symmetric, cylindrically symmetric, or planarly symmetric. In other words, to keep the math from getting overly complicated, you are responsible for solving equations with Gauss’ law only in situations which are highly symmetrical. Some examples of this are Inside and outside of a solid sphere made of conducting or insulating material. For this solution the Gaussian surface is a sphere. Inside and outside of a thin spherical shell. Again, the Gaussian surface is a sphere. An infinitely long, thin line of charges. The Gaussian surface for this is a cylinder which is colinear with the line of charges. An infinitely large, thin plane of charges, which we just did. Clearly you can use a Gaussian surface of a cylinder whose ends are parallel to the plane of charges. And 2 infinitely large, thin parallel planes of charges. Again, we just did this. You solve for a single plane of charges first and then do what we just did. Do not forget the three kinds of charge densities which you are responsible for being able to use with Gauss' law depending on whether the charge is one, two, or three dimensional. Linear charge density, surface charge density, and volumetric charge density. Actually, you know what, there is one more thing I want to review. Outside the surface of a uniformly charged sphere, the electric field is the same as if the charged sphere were a point particle. In order to prove this, let’s use an example of a solid, uniformly charged sphere with charge capital Q and radius lowercase a. Let’s create a Gaussian surface which is an imaginary sphere concentric with the charged sphere where the radius of the imaginary sphere is lowercase r and lowercase r is greater than the radius of the charged sphere, lowercase a. Bobby, please prove that this charged sphere, outside the surface of the charged sphere, has the same electric field as a point charge. Wow. We are going to do more? We’re like 29 minutes into this lecture already, right? So? I don’t care. I wanna keep learning! Yes Bo. We are going to continue to learn more. Bobby, please? Okay. According to Gauss’ law, the electric flux through the closed surface of the Gaussian sphere equals the integral over the Gaussian sphere of the electric field through the Gaussian sphere times the cosine of theta with respect to area. The direction of the electric field is always outward and the direction of every infinitesimal area dA is also always outward. The angle between those two directions is zero degrees. The cosine of zero degrees is one. The electric field is constant over the entire Gaussian sphere, so we can take that out from the integral. The integral of dA over the Gaussian sphere is just the area of the Gaussian sphere. The surface area of the Gaussian sphere equals four pi times the square of the radius of the Gaussian sphere which is lowercase r. The charge inside the Gaussian sphere is the charge on the charged sphere or capital Q. We derived that the electric field outside a charged sphere equals the inverse of four pi times the permittivity of free space all times the charge on the sphere divided by the square of the distance from the center of the sphere to the location of the electric field. Oh, but we can replace the parenthetical variables with the Coulomb constant because the Coulomb constant equals the inverse of four pi times the permittivity of free space. I just proved that, outside the surface of a uniformly charged sphere, the electric field is the same as if the charged sphere were a point particle. And, clearly, it works at any point outside the sphere, no matter how close to the surface. Thank you Bobby. Now, realize this is true of a conductor or an insulator, however, the electric field inside a conductor will be zero, and inside an insulator the electric field depends on the radius and charge distribution, and can be derived in a similar manner to what Bobby just did. We are not going to do that today because, well, we’ve been learning for quite a while today. So, that concludes my review of electric flux and Gauss’ law for AP Physic C: Electricity and Magnetism. Next time we are going to review electric potential. Thank you very much for learning with me today, I enjoy learning with you. So, what did you think? About what? Is Gauss’ law less tangible than mechanics? Oh, you mean, is Gauss’ law, which uses electric flux, which is a measure of the number of imaginary electric field lines which pass through an imaginary Gaussian surface and those imaginary field lines are caused by highly symmetric groups of stationary point charges which are imperceptible to the naked eye, is that less tangible than mechanics? Yeah, I guess that was my question. Yes. But I love it! I guess it just takes a bit of imagination to be able to visualize. Which is why we need to practice! Right. Yep.