We've done it. We now have one of Maxwell's equations in both forms. We have Gauss's Law in both the integral form, the integral of E dot dA is Q enclosed over Epsilon naught. This is the integral form that perhaps you're used to. Then we also showed through the use of the divergence theorem, that the integral of Del dot E is the divergence of the E field, is equal to Rho over Epsilon naught, where Rho needs to be a continuous charge distribution in space. This is the differential form, and they really are the same thing. You can see the relationship between them, I think, if you just think about a point charge or just a small charge. We have thought extensively about that, and when you have a point charge like this, we'll go back to field lines. For now, our field line is coming out. In these kinds of problems, the kinds we do in freshman physics, it makes sense to use the integral form, and the reason is, you can use it with just the charge enclosed. You can have a Gaussian surface, you can do your vector calculus out here, and you can do it with finite charges. This can be a point charge, this can be a sphere of charge, this can be a rod of charge, it doesn't matter. As long as it's enclosed, you can just stick all of Q right there, and you can do your vector calculus separate. The differential form isn't as useful here, because you really can't use it for a point charge here. Your charge density has to be described by a smooth function. You don't want to get into where the charge density is one value in the sphere and then it's 0 outside, and they forget to do those boundaries differently for the E field, it's all very confusing. That's the reason you really work with the integral form more, is because it's more useful in these kinds of situations. The differential form is useful really when you're manipulating fields and thinking about fields. Perhaps if Rho is zero, it's really telling you how the E field behaves. It behaves in a way that it only diverges when a charge shows up, and that's another thing you can see. We talked about how divergence is the creation of a field. Well, the E field is created when there's charge, so the field is very divergent right here where there's a charge. The field is not divergent out here. It's just field lines going right through. [inaudible] the question, can you use it at all? Was it for anything other than manipulating fields? I want to show you a problem where you can use it. We'll do one problem and do it one way with integral form, and we'll do it back with the differential form just to show you it is possible. Let's look at just a charged sphere with charge density Rho, constant throughout the sphere, and just ask what is E inside? If you're going to use the integral form, you would draw a spherical Gaussian surface inside centered on the sphere, all the standard stuff we talked about in the Gauss's law chapter, and you'd say the integral E dot dA, well, E vector and dA vector are the same direction, so the dot product goes away, E is constant over the surface due to symmetry, it comes out of the integral, and you end up with E times the area of that Gaussian surface, and it's some radius, r, is four Pi r squared equals the charge enclosed. If this thing has a uniform charge density Rho and the charge enclosed is Rho times the volume of the Gaussian sphere, Rho times 4/3 Pi r cubed over Epsilon naught. All kinds of stuff cancels, and what you find is that inside a sphere like this, E or the magnitude of the E field we're finding is, what is it? Rho r over three Epsilon naught. The E field inside a sphere is 0 right at the center, has to be due to symmetry, and it just increases linearly with radius. The E field gets bigger, pointing out in a radial direction. We figured out that out from the symmetry, so we can give it a vector description. That's what the E field does. Now with the differential form, you can go the other way. You could try to take the divergence of that E field and see if you get the same charge density. That's one thing you could do with it. Let's do that, see what happens. Well, it's a little complicated. This is Gauss's law. This is always true. Regardless of what coordinate system we're using, Del dot E is Rho over Epsilon naught. But how you write Del dot E mathematically doesn't depend on the coordinate system. In Cartesian coordinates it was just the partial d dx of the Ex of the function in front of the I hat, and d dy of the part in front of the J hat, d dz in front of the k hat. Well, if you go into cylindrical and spherical coordinates, the unit vectors change direction depending on where you are. That's the evil thing about cylindrical and spherical coordinates, and that makes the vector derivatives, the divergence and the curl very complicated. Let me show you then what the spherical divergence looks like. If we were to have Del dot V, and when we go back to V, let me make it clear, I'm not talking about the electric field here. This is just the spherical divergence for any vector field. It's one over r squared d, dr of r squared times Vr. That's one term. Remember, the divergence gives you a scalar, so you don't have to worry about I hat j hat and k hat, this is just the part Vr, is the part that would have gone in front of Vr hat if we were to write a vector field, Vr. Then it's one over r sine Theta, partial d, d Theta of sine Theta V Theta, that for one angle, and then there's a third term for third angle plus 1 over r sine Theta, and then it's dV Phi, d Phi. Those are the three terms. In Cartesian, you had an x part, there's the xVr, you had a y and you had a z. It's just much, much more complicated. But let's try to apply it. Now let's apply this big nasty thing to Del dot E. Well, fortunately, we have very nice symmetry that's going to help us out. Our E field only has the r hat component. We only have Er in this case. Our E Theta is zero. It never points in the Theta direction. Our E Phi is zero, it never points in the Phi direction. We do have it down already to one over r squared d dr of r squared, and then this is Rho r over three Epsilon naught. Rho r over 3 Epsilon naught. There's our divergence. Just running on a board here, so I'll take some shortcuts. This becomes Rho r cubed over 3 Epsilon naught, and I take the derivative of that, the partial with respect to r, that's 3 Rho r squared over 3 Epsilon naught, and then times 1 over r squared, so 1 over r squared times 3 Rho r squared over 3 Epsilon naught. Now you can see r squared cancel, the 3s cancel, and you end up with the constant Rho over Epsilon naught, which of course we know is what Del dot E is, that's Gauss's law. That was one problem example where you can go both ways, you go the opposite direction, but you're going apply both to the same problem. If you wanted to make it more interesting, you could do instead of a constant Rho, you could do a Rho that varies in space, and then you could do this way, and you can get a weird E field and go that way, and you could find a function for Rho instead of just finding Rho over Epsilon naught. Anyway, they are the same law. Integral form, differential form, they tell you the same thing, they're just useful at different times when you're working on problems.
