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Doc Al

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thanks! my book just says "relates" not proportional, i like your wording better!

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Flux is determined by taking a surface integral over a vector field. That is, you want to find the total amount of whatever going through a surface, so you find the rate at which stuff moves through the surface at every infinitesimal point and add them all up.

First consider a sphere with the charged particle at the center. Since every point on the surface of the sphere is the same distance away from the particle, the field on the surface is the same throughout (E is only a function of radius). Therefore, the flux equals the electric field at the surface times the surface area.

(for a sphere) [tex]\Phi = (\frac{kQ}{R^2})(4\pi R^2)=4kQ\pi[/tex] which is a constant.

(4*Pi*R^2 is the surface area of a sphere).

This means that no matter what the radius is, the flux is the same.

That solves the sphere cases, but the cubes require a bit more reasoning (I'm trying not to get too caught up in the math).

Visualize this: The charge particle is at the center of a small sphere. There is a cube that surrounds that sphere. Then there is another sphere that surrounds the cube, centered at the charged particle. Now, you know intuitively that the field falls off with distance. Therefore, it makes no sense that the cube should have a greater flux than the small sphere that it encloses, because the smaller sphere has the same flux as the larger sphere (as shown from above). You can sort of use the same argument for why the cube can't have a smaller flux than the spheres. This only works, however, if the cube is closed so none of the field 'leaks' out.

If you want to understand this a bit more rigorously, you can read about conservative fields and the derivation of Gauss' law.

I hope that helps.

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thanks for the explanation!

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James R

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1. Make the field strength stronger, so the field lines become more dense (more lines in a given volume); or

2. Increase the size of the area intercepted by the field lines; or

3. Alter the orientation of the area so it is "more perpendicular" to the field direction at the surface.

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Chronos

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