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Gauss's Divergance Theorem and Stokes's Theorem |
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| Dec26-03, 11:27 AM | #1 |
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Gauss's Divergance Theorem and Stokes's Theorem
I've been reading the text: Electricity and Magnetism, by Purcell.
I understand about the integral forms describing the electric field, but when trying to answer questions at the end of the chapter on Gauss's and Stokes's therorems I have not been able two. These two theorems supposedly transform the integral equations that describe an electric field into differential equations, yet there are still integrals in the equations: Gauss: [tex]Integrate[F*da,over surface]=Integrate[div F*dv,over volume][/tex] [tex]div F=4\pi\rho, del^2*potential=-4\pi\rho[/tex] Stokes: [tex]Integrate[F*ds,over circ.]=Integrate[curl F*da,over surface][/tex] [tex]del X A[/tex] Questions: 1. I'm having trouble with looking at field lines and judging if div F=0 (or not zero) or curl F=0 (or not zero) 2. How to calculate flux/volume for Gauss problems. I see in the book they use the midpoint. Or flux/area for Stokes problems. Why do the choose the midpoint? 3. Also would someone explain the sidways derivative of the curl, I don't understand why it is the way it is. |
| Dec30-03, 08:47 PM | #2 |
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Now imagine an aquaduct where the water flows faster the further it is from the bank you're standing on. As you move perpendicular to the flow direction, the flow rate gets faster. This flow has a nonzero curl. Place a stick perpendicular to the flow, the far edge wants to move faster than the near end making the stick want to turn. Now look at the math for curl. If the flow is in the x direction then the curl is given by the derivatives dv/dy and dv/dz. That is the velocity v is changing when you move perpendicular to the flow direction. |
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