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FS98

- 105

- 4

If you use a Gaussian surface that encloses no charge you find that the electric field is equal to 0. But if there is a charge outside of that Gaussian surface, it is not true that the electric field is 0 on the Gaussian sphere.

I want to say that the electric field calculated will only give the electric field due to the enclosed charges, but I don’t think that’s true either. If we solve for the electric field of an infinite line of charge, the charges on the outside of the Gaussian surface do have an impact on the calculation. If the line of charge we finite, we would have the same enclosed charge, but gauss’s law wouldn’t work so easily because there would no longer be the same symmetry?

So what exactly does the electric field calculated by gauss’s tell us?