![]() But obviously there must be one that does. Of course, this intuitive approach is quite weak, with no "proof." And it probably doesn't work for many other problems. ![]() But positive z direction is out of the parabola. Thus, q(enclosed) vanishes for any surface inside the chunk of conductor, which tells you there are no net charges anywhere in there. We used ∂v/∂t × ∂v/∂s, so t cross s is x cross y which is in the positive z direction. As we see, t is equivalent to x, and s to y. We need a systematic procedure to find which way the vector really points (and maybe the expression we get using this procedure contains a and b, so we can generalize without having to get an absolute answer).Īs for this problem, I did find a way around plugging in values, with intuition. In this case, as a and b change, the directions of everything change, so a plug-in at one value doesn't work for a plug-in for another. So maybe v(t,s) =, and we want to see how flux changes as we vary a and b. Sometimes we might not know what the surface looks like so we cannot just observe where the vector points and say, "Oh, that's outward." Also, we might be solving abstractly. Is there some better way to find out where the normal vector points? Other than by plugging in some value. It might be an electric field, and then perhaps there is no net charge inside the parabola. ![]() It might be heat transfer, and then there is no net heat entering from the sides (we still don't know about the bottom). If it also equals zero, then yes, the mass of fluid inside is constant.īut if the vector field doesn't represent a fluid's velocity, it might mean something else. To make sure, you would need to compute the flux of fluid through that region of plane. There might be net fluid escaping or entering through that plane. If you just seal the "hole" with a flat plane, then, no, you cannot be sure yet that mass inside is constant. Compute the electric force between two charges of 5. There is a "hole" on the bottom of the parabolic surface. Scroll down the page for more examples and solutions on how to use the formula. In pictorial form, this electric field is shown as a dot, the charge, radiating lines of flux. I mean, what region? The surface is not closed. In words: Gauss’s law states that the net electric flux through any hypothetical closed surface is equal to 1/0 times the net electric charge within that closed surface. However, we cannot assume that the mass of fluid inside the region is unchanging yet. formula is substituted for the true catenary, it being argued. If the vector field is really velocity of fluid (I think it is in this problem), then the mass of fluid going into the "region" is exactly equal to the mass of fluid going out. Electric Pole and Tower Lines feature is the large number of specifications inserted at Field.
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