Example 2- Electric field of a uniformly charged spherical shell
As in another example to Gauss’s law, let’s try to calculate the electric field of a spherical shell charge distribution. In this case we have a spherical shell object, and let’s assume that the charge is distributed along the surface of the shell. It’s like basketball for example. It’s charged with certain Coulombs along its surface uniformly. We’d like to calculate the electric field that it generates at different regions. So here’s our spherical shell. Let’s assume that it’s positively charged to some Q Coulombs and its radius is big R.
First, we’d like to calculate the electric field inside of this spherical shell. In other words, our point of interest is somewhere over here at an arbitrary location, or let’s say some little r distance away from the center such that little r is smaller than big R inside the region and we choose our Gaussian surface, closed hypothetical surface, in the form of a sphere such that it passes through the point of interest.
Gauss’s law says that integral of E dot dA over this closed surface, let’s denote the surface as closed surface S1, is equal to q-enclosed over ε0. Before we deal with the left-hand side of the Gauss’s law in this case, let’s just look at the right-hand side. Right-hand side is telling us that we have q-enclosed and that is the net charge enclosed inside of the volume surrounded by the Gaussian surface, in this case the Gaussian sphere. That is a surface that we choose. In other words, we are talking about this region.
When we look at that region, we don’t see any charge over there. The whole charge is distributed along the surface of the spherical shell. There’s no charge inside. Therefore, q-enclosed is 0. Since q-enclosed is 0, therefore we can say that the electric field inside of the spherical shell is 0. No source, no charge.
For the outside region, electric field for little r is larger than big R. In that case, our point of interest is somewhere outside. Again, its position relative to the center is given by little r. Using the symmetry of the distribution, we will again choose a spherical closed surface such that it is passing through the point of interest. Something like this. When we look at the surface, the electric field generated by our source is going to be pointing radially outward direction at the point of interest. If we go to the other points along this surface, again, we will see that electric field will be radially out and so on and so forth along this surface.
Since as long as we are along this surface, let’s call that surface as S2, we will be same distance away from the charge distribution, then the magnitude of the electric field along this surface will be constant. Furthermore, if we just look at incremental surfaces at different locations on this sphere S2, we will see that the area vector will be perpendicular to those surfaces and since we’re talking about a spherical surface, these dA‘s will also be in radially outward directions.
Therefore the figure shows us that wherever we go along this surface of sphere S2, the angle between electric field vector and the area vector will always be 0 degrees. In explicit form therefore, the Gauss’s law which is E magnitude dA magnitude times cosine of the angle between E and dA, and it is 0 integrated over surface S2, will be equal to q-enclosed over ε0. Cosine of 0 is just 1. Electric field is constant over this surface, we can take it outside of the integral. That leaves us electric field times integral over surface S2 of dA is equal to q-enclosed over ε0. Integral of dA over surface S2 will give us the surface area of sphere S2, which will be 4π, little r2, times the electric field will be equal to q-enclosed.
Now again, we go back to, in this case, look at the region surrounded by sphere S2. That is this region. When we look at that region, we see that the whole charge, which is distributed along the spherical shell, is inside of the region surrounded by surface S2. The amount of charge along that spherical shell is Q, therefore q-enclosed is equal to big Q. Then we will have Q over ε0 on the right-hand side. From here, leaving electric field alone, we will end up with Q over 4π ε0 r2.
Again, writing down this one in vector form, we will multiply it by the unit vector in radial direction because the electric field is pointing radially outward. Furthermore, again, when we look at this expression, it’s a familiar expression. It is basically equal to the electric field of a point charge. So when we look at this distribution then, for the points outside of the distribution, in other words, for the exterior points, then the spherical shell distribution is behaving like a point charge.
The electric field that it generates is equal to the electric field of a point charge. In other words, it is behaving as if its whole charge is concentrated at its center. So the system, in a way, becomes equivalent to as if I have a positive test charge with a Q coulombs of charge and I’m interested in its electric field some r distance away from the charge. That electric field will be radially out and it will have exactly this magnitude.