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[tex]\nabla\cdot E = \frac{\rho}{\epsilon}[/tex]

Loooking at this it seems like you would have no divergence except on the suface. How does that help?

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- Thread starter Swapnil
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- #1

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[tex]\nabla\cdot E = \frac{\rho}{\epsilon}[/tex]

Loooking at this it seems like you would have no divergence except on the suface. How does that help?

- #2

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More practically, given a choice between expressing concepts as integral equations or expressing them as differential equations, it is usually easier to express things as differential equations. This is especially true in cases where problems reduce to one or two dimensions. For example, imagine that you're trying to find the capacitance of a simple parallel plate capacitor. If you were to do this with integrals, you'd have to integrate out to infinity in two dimensions, and it wouldn't be easy. With the differential equation, you can easily see that the electric field between the plates is constant with position, and you can apply the boundary conditions quite rapidly.

- #3

Claude Bile

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Having divE = 0 is a very useful constraint when it comes to studying electromagnetic wave propagation (among other things).[tex]\nabla\cdot E = \frac{\rho}{\epsilon}[/tex]

Loooking at this it seems like you would have no divergence except on the suface. How does that help?

Claude.

- #4

vanesch

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Together, the divergence and curl of a vector field uniquely describes it. (In mathematics, this is known as Helmholtz's theorem.) As a result, we can express the electric and magnetic fields in terms of their divergences and curls, and this is precisely what Maxwell's equations do.

If I'm allowed to nitpick: the divergence and curl of a vector field uniquely describe it on the condition that they vanish at infinity. Because the condition of vanishing divergence and curl is nothing else but the condition that the potential is harmonic. All wave solutions are of that kind (but they don't vanish at infinity for all times).

- #5

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In a fluid, to say that divergence is positive, means that the fluid, say a gas, is expanding at that point. To say that the divergence is negative means that the fluid is compressing at that point.

Since gases are referred to as compressable, and liquids incompressable, usually for a liquid you will set divergence to zero. However even liquids, and solids, have slight amounts of compressability, but it's usually negligable.

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The divergence operator gives us the information about how the field is changing inside of a closed surface. By other words, if there is field being "created" or if there is field being "destroyed". You can easily understand this by looking at the Gauss Theorem. If the flux, throught a closed surface, change then this means that the field is changing inside the surface and hence the divergence will be non-null somewhere inside the surface. If the flux is keep unchanged (or by other words, if everything that get inside from one side of the surface get outside by the other side of the surface) then this means that there are no sources inside and so the divergence will be zero somewhere inside the surface.

So, as you can see, the divergence operator is very usefull when working with vector fields.

I supose that you now understand why this happens.

So, as you can see, the divergence operator is very usefull when working with vector fields.

I mean, for point charges, the divergence is zero everywhere except where the charge is located.

I supose that you now understand why this happens.

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