A charge of negative 30 microcoulombs is distributed uniformly over the surface of a spherical volume of radius 10.0 centimeters. Determine the electric field due to this charge at a distance of 2.0 centimeters from the center of the sphere. Determine the electric field due to this charge at a distance of 5.0 centimeters from the center of the sphere. Determine the electric field due to this charge at a distance of 20.0 centimeters from the center of the sphere.
In this situation, we have a given amount of charge that is distributed uniformly over the surface of a sphere. And the sphere’s radius is given to us as well. We want to solve for the electric field that’s caused by this uniformly distributed charge at various distances from the center of the sphere that they rest on. Since the three questions that were asked are identical except for the distance from the center of the sphere at which we’re to solve for the field, let’s keep a tally of these three different distances off to the side and start with a sketch of this spherical object.
Here are these three distances from the center of the sphere. We’ve called them 𝑟 one, 𝑟 two, and 𝑟 three as well as capital 𝑅, which is the radius of the sphere itself 10.0 centimeters. In addition to all this, there’s a net negative charge uniformly distributed over the surface of the sphere. And we’re given that charge amount: negative 30 microcoulombs. We can label that charge capital 𝑄. For our first question, we want to solve for the electric field created by this charge 𝑄 at a point 𝑟 one from the center of our sphere.
Knowing that 𝑟 one is 2.0 centimeters, we can imagine it looks something like this within the sphere’s volume. When we consider the electric field at 𝑟 one created by charge 𝑄, which is distributed uniformly over the surface, we can start to imagine how the electric field lines from each individual charge on the surface interact with the field lines of the other charges. For example, let’s consider the electric field lines that come to this particular charged we’ve marked out on the surface. It’s a negative charge, so we know the field lines will come toward it rather than away from it. And these field lines might look something like this.
And looking at them, we see that they do indeed have an effect at the point we’ve labeled 𝑟 one. In other words, if it was just this charge on the surface of our spherical volume, then the electric field at this point would be nonzero. The field affects that point, but let’s keep looking at the other charges on the surface. In particular, let’s consider the charge on the spherical surface that opposite the first one we chose. We can draw the electric field lines for this charge as well. They would look something like this, identical to the field lines from the first charge we picked. In fact, if we went around this circle, one by one drawing in the electric field lines from each of the charges on the sphere, then we would see all these electric field lines overlapping and in fact cancelling one another out.
As an example of that, if we consider the point at the very center of our sphere, then the electric field created by the first charge we’ve drawn field lines for and the field created by the second charge we’ve drawn field lines for oppose each other at that point perfectly. So just due to the effects of these two charges opposite one another on our sphere, the electric field at this point would be zero. And when we consider not just one pair of charges opposite one another on the circle but all pairs of opposite charges, its effect is reinforced. Here is the net effect of all this. It’s a surprising result, but it has to do with the fact that our charge is uniformly distributed over this surface.
The overall effect is that inside the sphere the electric field lines from the various charges on the sphere’s surface perfectly cancel one another out at every location, not just at the center point, but everywhere within the sphere the electric field is zero. And again this has to do with the electric field lines of this uniformly distributed charge working against one another to cancel each other out. So what does all this mean for the electric field at 𝑟 one? Well since 𝑟 one is inside the sphere, that means that the electric field there is zero. And in fact, we can say even more. The electric field magnitude at 𝑟 two, which is a distance of 5.0 centimeters, from our 10.0 centimeters radius sphere is also inside the sphere and therefore also equal to zero.
As we’ve seen, because of the way the electric field lines of the charges on the sphere surface work against one another, the electric field at any point within this volume is zero. But take a look at this! 𝑟 three our final radius value is 20.0 centimeters, which is indeed outside our spherical object. We could say that 𝑟 three is a point some distance such as this away from our sphere’s surface. And what we’d like to know is what’s the electric field there at that point. To figure this out, we’ll once again use the fact that the charge on the spherical volume is distributed uniformly. That comes in handy once again! Because that uniform distribution, that means we can simulate the location of all of this charge 𝑄 at a location at the center of the sphere.
So when we’re calculating the electric field at a point somewhere outside the sphere, we can effectively concentrate all of the charge that the sphere has on its surface at a point in its center for the purposes of this calculation. So we imagine the full amount of charge 𝑄, negative 30 microcoulombs, moves from the surface to the center of this volume. And again this isn’t happening physically, but from a mathematical perspective this will help our calculation. Now we effectively have a point charge 𝑄. And a known distance away, we want to solve for the electric field created by this point charge. That scenario may sound familiar. And we can recall that the electric field vector created by a point charge 𝑄 is equal to that charge 𝑄 multiplied by Coulomb’s constant 𝑘 all divided by the distance between the charge in the point at which we’re solving for the field.
And since this is a vector were calculating, it has a direction either away from or towards the point charge we’re considering. In our case then, we can say that the electric field at 𝑟 three is equal to Coulomb’s constant times our overall charge 𝑄 divided by 𝑟 sub three squared in the radial direction, where that direction is given along a line between our point charge and the point we’re calculating the field. Looking at this equation, we already are given 𝑟 sub three, and we know 𝑄. The only thing we still need to know is Coulomb’s constant 𝑘. When we look up or recall that constant, we see it’s equal to 8.99 times 10 to the ninth farads per meter.
So we enter in this value for 𝑘, we write our charge 𝑄 as negative 30 times 10 to the negative sixth coulombs, and we rewrite our radial distance of 20.0 centimeters as 0.20 meters. When we calculate all this, we find a result to two significant figures of negative 6.7 times 10 to the sixth in the 𝑟-hat direction newtons per coulomb. That’s the electric field at point 𝑟 three. And note that this field is a vector with both magnitude and radial direction.