A total charge of 5.6 times 10 to the negative sixth coulombs is distributed uniformly throughout a cubical volume whose sides are 12 centimeters long. What is the charge density within the cube? What is the electric flux through a cube that is concentric with the charge distribution and that has 12.0-centimeter long sides. What is the electric flux through a cube that is concentric with the charge distribution and that has 15.0- centimeter long sides? What is the electric flux through a spherical surface of radius 2.5 centimeters that is also concentric with the charge distribution?
Lots of questions here! Let’s start out with the first one: what is the charge density within a cube? To start off, we’ll sketch out what this charge distribution looks like. We’re told this amount of charge, 5.6 times 10 to the negative sixth coulombs, is distributed uniformly over a cubical volume. Here we’ve drawn one face of the cube with side lengths of 12 centimeters, and we imagine the cube goes just this far into or out of the page.
Given this distribution, we want to know what is the charge density within this cube. That term charge density is a little bit like the regular term for density. In fact, we can use the same symbol 𝜌 to represent charge density, which instead of being a mass per unit volume is a charge per unit volume. To solve for the charge density within the cube then, we’ll want to know just how much charge is in it and also just how much volume it takes up. We’re given that the total charge involved is 5.6 times 10 to the negative sixth coulombs.
And when it comes to volume, since we’re working with a cube, we know the volume is equal to the length of one of the sides of the cube cubed. We know that 𝐿, the length of each side of this cube, is 12 centimeters. But when we plug in for this length, we’ll want to use that SI base unit of length, which is not centimeters but meters. That means we’ll take our length of 12 centimeters and rewrite it as 0.12 meters. And now that our units are set up properly, we’re ready to calculate charge density.
To two significant figures, we find it’s 3.2 times 10 to the negative third coulombs per cubic meter. That’s the density of charge in this distribution of charge. Next, considering this distribution, imagine that we surround it with a virtual surface, a cube of side length which is also equal to 12 centimeters. And we know that this virtual surface is a closed surface; it’s complete; there are no holes in it. Based on this, our next question asks what is the electric flux — we will call it 𝜑 sub 12 cm — through this virtual surface.
In other words, as the electric fields created by each one of the charges in our charge distribution emanate outward through the closed surface, just how much electric flux we wonder is there. To figure this out, we’ll be helped by a law which is known as Gauss’s law. This law says that the electric flux through a closed surface, like the one we have here, is equal to the charge enclosed by that surface all divided by the permittivity of free space, 𝜀 naught.
Here’s how we can picture Gauss’s law. Say we have some charge and an irregular surface that completely surrounds and encloses that charge. Assuming this point charge is positive, we know it will create electric field lines moving out through this enclosing surface. If we wanted to calculate the electric flux through this surface, one way to do it is to find the electric field at each segment of the closed surface area multiply it by that area element and then multiply that by the cosine of the angle between the field and the normal to that area segment.
We can see how this would quickly become very complicated for irregularly shaped surfaces, like the one we have here. Gauss’s law simplifies all this. It says to solve for that electric flux, we don’t need to know the electric field or the area or the cosine of the angle between them. All we need to know is the charge enclosed by our closed surface. Once we know that, we can use it along with 𝜀 naught to calculate the electric flux.
Now in our scenario, we don’t have a particularly complicated surface, but we do have a complicated charge distribution. To calculate electric flux using 𝐸 times 𝐴 times the cosine of 𝜃, we would somehow need to calculate the electric field from each one of the charges in this distribution at each point on our cube. Or we could calculate the charge enclosed by our cube and divide that by 𝜀 naught. Let’s go that route.
As we do so, we recognise that 𝜀 naught is a constant value, which we can treat as 8.85 times 10 to the negative 12th farads per meter. Knowing that, the next step in solving for 𝜑 sub 12 cm is to figure out just how much charge is enclosed by our virtual closed surface. We’re told that our closed surface has side lengths which are the same as the side lengths of our charge distribution and that the two are concentric; that is, their centres are the same. This means that our virtual surface completely overlaps and encloses all of the charge distributed over this cube.
That means that 𝑄 enclosed is equal to our total charge, 5.6 times 10 to the negative sixth coulombs. And then plugging in for 𝜀 naught, we’re now ready to solve for 𝜑 sub 12 cm, the electric flux through this surface of 12- centimeter side length. The result is 6.3 times 10 to the fifth newton-meters squared per coulomb. And see how much easier it was to calculate this flux using Gauss’s law! For our next part of the problem then, we imagine that our virtual surface changes. Now instead of having side lengths of 12 centimeters like our charge distribution, our virtual surface expands to have side lengths of 15 centimeters.
The question now becomes what is the electric flux through this new expanded surface. We’ll call this 𝜑 sub 15 cm. And to solve for it, let’s again consider Gauss’s law. Recall that this law shows us that the only thing we need to calculate the electric flux over a closed surface is to know just how much charge is inside the surface. That and the constant value 𝜀 naught give us flux. Gauss’s law says nothing about the size or shape of the enclosing surface itself. All that matters is how much charge is inside.
Considering our expanded virtual surface, we can say that nothing about the charge it encloses has changed. And therefore, according to Gauss’s law, the electric flux through this surface will be the same as that through the 12-centimeter surface. In other words, just as much electric flux would pass through the walls of this 15- centimeter side length cube as passed through the walls of the 12-centimeter side length cube that was also centred on the charge distribution. So our previous answer for electric flux applies here too: there are 6.3 times 10 to the fifth newton-meter squared per coulomb of electric flux passing through our virtual cube of side length 15 centimeters.
And in fact, no matter how big our virtual surface got, as long as it still encloses the same amount of charge, the flux through it would be the same. Now for the last part of this question, we want to consider a different type of virtual surface entirely. Now instead of a virtual cube, we’re considering a virtual sphere centred on our charge distribution. This sphere we’re told has a radius of 2.5 centimeters, and we want to calculate the electric flux through it. We’ll call this 𝜑 sub s.
Applying Gauss’s law once more, we’ll say that 𝜑 sub s, the electric flux through this virtual spherical surface, is equal to the amount of charge it encloses divided by 𝜀 naught. Looking at our sketch, we can see that our virtual surface no longer encloses all of our charge but rather just some fraction of it. It’s that fraction we’ll want to calculate in order to be able to solve for 𝑄 enclosed. Let’s consider how this works.
The charge enclosed by our sphere is going to be equal to the total charge, we’ll call that 𝑄, multiplied by a ratio of volumes. We’ll have the volume of our sphere in the numerator and the volume of our cube in the denominator. Whatever this fraction is, this ratio of volumes, when we multiply it by our total amount of possible charge, 5.6 times 10 to the negative sixth coulombs, that will be the charge enclosed by our virtual sphere. So what is 𝑉 sub s, the volume of our virtual sphere, and what is 𝑉 sub c, the volume of our cube of charge?
We can recall that in general the volume of a sphere is equal to four-thirds 𝜋 times the radius of the sphere cubed. And our radius is given as 2.5 centimeters. So we can write that expression in in the numerator for our volume fraction. Next, the volume of the cube of our distributed charge is equal to the length of the side of that cube cubed. That length as we know is 12 centimeters, so we insert that for the length of the side of the cube.
Note that we’ve kept our length units as centimeters, though usually we convert to meters. The reason we can get away with this is that these lengths units will cancel out, cubic centimeters on top and on bottom. So for simplicity, we’ll leave them as they are. Now to calculate the charge enclosed by our virtual sphere, all we need to do is to plug in for 𝑄. That’s the total charge in this distribution. We now have an expression for the charge enclosed, and we know that if we divide that value by 𝜀 naught, then that fraction will give us 𝜑 sub s, the electric flux through our virtual sphere.
So 𝜑 sub s is equal to this big expression. Notice, as we pointed out, that the units of cubic centimeters cancel. And when we calculate all this, we find a result of 24 times 10 to the third newton-meters squared per coulomb. This is how much electrical flux passes through our virtual sphere placed in this charge distribution.