### Video Transcript

A very long, thin wire has a uniform linear charge density of 50 microcoulombs per meter. What is the magnitude of the electric field 2.0 centimeters from the axis of the wire?

When we sketch out this wire, we’re told that it’s a very long, very thin wire. We can assume it goes very far beyond what we’ve drawn here up and down. And this wire carries a linear charge density. We’ll call that charge density 𝜆. Knowing all this, the question asks if we pick a point 2.0 centimeters away from the axis of the wire, what is the magnitude of the electric field there?

Now, the first thing to notice is that it doesn’t matter which direction from the wire we go. As long as we’re moving 2.0 centimeters out from its axis, the electric field will be the same everywhere at that distance away. To solve for the magnitude of the electric field at this distance away from the axis of the wire, we can recall the law that’s known as Gauss’s law. This law says that if we create an imaginary closed surface of volume, then the charge enclosed in that volume, 𝑄 enclosed divided by the permittivity of free space 𝜖 naught, is equal to the electric flux through that closed surface. That is the integral of the electric field times the area of the surface.

Since applying Gauss’s law to a particular scenario, like our very long, very thin wire, always involves creating an imaginary enclosed surface, let’s figure out what that surface will be in our case. Since our wire is really symmetric, we may as well make a cylinder around this wire, be our imaginary Gaussian surface. We’ll say that our cylinder has a radius, which is as big as the distance away from the wire that we want to solve for the electric field. And we’ll say that our cylinder has a height ℎ. And notice that the cylinder just barely encloses the point of interest 2.0 centimeters off the wire axis. As we go along, we keep in mind that this indeed is an imaginary surface. It doesn’t really exist. But we’re using it because it’s convenient for helping us apply Gauss’s law to the scenario. Recalling that law, it involves the charge enclosed in our imaginary surface. So let’s solve for that amount of charge now.

What we have is a long thin wire with linear charge density that we’ve called 𝜆. We can see that 𝜆 is an amount of charge per unit length. So the question is “how much of the wire is enclosed in our imaginary cylinder?” See that we’ve called the height of our cylinder ℎ. That means that a distance of ℎ along our wire is enclosed in our imaginary shape. That means that if we take our linear charge density 𝜆 and multiply it by the distance ℎ, then that will be the amount of charge enclosed in our imaginary surface. Since this quantity, 𝜆 times ℎ, is equal to 𝑄 enclosed in our Gauss’s law equation, we know that the next step is to divide this by 𝜖 naught, the permittivity of free space. And then, we can start to consider the other side of Gauss’s law, the integral of the electric field over the area of our surface.

To start to get a sense for the geometry of this right side of our equation, and indeed it does involve geometry because there’s a dot product, let’s sketch out once more the direction of the electric field lines coming from our long thin wire. Since our wire has a positive linear charge density, that means the electric field lines will come off of the wire rather than pointing to it. As we saw earlier, those electric field lines move out radially from this wire. So if we looked down on it from above, it would look a bit like the spokes on a bicycle wheel.

This has implications for the dot product we’ll take between 𝐸 and 𝑑𝑎. Because the electric field lines always move radially outward from our long wire, that means that if we were to draw a little box representing a differential area element on the surface of our cylinder, then the electric field lines coming from our wire that would go through that area element would be pointed straight out through it, straight away from the wire. And that’s interesting, because notice that the area vector of this area element we’ve drawn, this little rectangle, also points straight out from the wire. That is, the area vector and the electric field point in the same direction.

This fact that the electric field and the area element both have the same direction to them has implications for how we’ll do this integral in our case. It means we can write that integral simply as 𝐸 multiplied by 𝑑𝑎, no vectors this time, integrated over the area of our surface.

Now there are two other interesting things we can notice here as we can consider this area integral. First, as we said, the electric field lines move radially outward from our thin wire, which means that if we consider the top and the bottom surfaces of our cylinder, that is, the parts of this closed surface that are circles, then we can say that no electric field lines pass through those parts of the surface. If any electric field lines did pass through those parts of the surface, that would mean that they’re no longer traveling radially outward from the wire.

This means that when we calculate our area integral, we don’t need to include these top and bottom parts of the surface. That’s because the flux through them is zero. When we calculate our area integral then, we only need to consider the band that’s around the wire. The second thing to consider here is the electric field magnitude at any location along this circular band. Since this band is always the same distance away from the wire access, that means the field strength on the band is always the same. In other words, as we integrate over the area of this band, the electric field will be constant at each point. And what can we do with a constant that’s inside an integral sign? We can factor it out.

We’re just about ready to calculate this integral. And in doing so, solve for the electric field magnitude at distance away from the axis of the wire. We’re told in the problem statement that that distance is 2.0 centimeters. For the purposes of our calculation though, let’s give it a symbolic name. Let’s call this distance 𝑟.

Labelling that distance as 𝑟, that means that the integral of the area over our imaginary shape, that is its total area where — recall once more that we’re not including the areas of the top and bottom of the shape since the flux through them is zero. This area then, the area of our circular loop, is equal to the circumference of the loop, which is two times 𝜋 times its radius 𝑟, multiplied by the height of our cylinder, what we’ve called ℎ. So then, 𝜆, the linear charge density of our wire multiplied by the height ℎ of our imaginary cylinder divided by 𝜖 naught, is equal to the electric field magnitude multiplied by two times 𝜋 times 𝑟, the radius of our imaginary cylinder, times its height.

And notice that the height of our imaginary cylinder ℎ appears on both sides of the equation and therefore cancels out. That’s a good thing because remember our problem statement has nothing to do with this imaginary cylinder. It was completely invented by us to help us solve the problem.

We can now see that, to calculate the electric field magnitude, we need only to divide both sides of the equation by two times 𝜋 times 𝑟. And now that we have an equation for that field magnitude, it’s just a matter of plugging in for 𝜆, 𝑟, and 𝜖 naught and then calculating this value. The one constant in all this, 𝜖 naught, is equal to approximately 8.85 times 10 to the negative 12th farads per meter. 𝜆, the linear charge density, we’re told is 50 microcoulombs per meter. And 𝑟, the distance from the axis of the wire we’re interested in the field strength, is 2.0 centimeters.

When we plug these values in for 𝜆, we write a value of 50 times 10 to the negative sixth coulombs per meter. Then for the radius 𝑟, we convert it from 2.0 centimeters to 0.02 meters. That’s so our distance unit, now meters, matches up with the distance units in the rest of our terms.

When we go to calculate this fraction, to two significant figures, we find a value of 4.5 times 10 of the seventh newtons per coulomb. That’s the electric field magnitude, the given distance away from the axis of this uniformly charged wire.