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The book says it is $E_0\pi r^2$ because the flux through the circle is equal to the curved part of the paraboloid.

I don't understand this, shouldn't the total flux be 0 for the whole surface? IN fact, since the E field is constant, then the flux must also be 0.

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up vote 9 down vote accepted

Be careful here.

Gauss's law tells you that the flux through the (whole) closed surface is proportional to the enclosed charge and therefore zero in this case.

That's one fact.

The second fact is that you have a constant electric field in this region of space, and that means that the flux through the circular end-cap (which is not a closed surface) is $E_0\pi r^2$.

Now we put the two facts together, the combination of the end-cap plus the parabaloid is a closed surface, which means that (because the flux through the end cap is pointed in and is therefore negative) we get

$$ 0 = \mathcal{F}_\text{end-cap} + \mathcal{F}_\text{parabaloid} = - E_0 \pi r^2 + \mathcal{F}_\text{parabaloid} $$

or, if we re-arrange things

$$ \mathcal{F}_\text{parabaloid} = E_0 \pi r^2 . $$

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For some reason I thought the "closed paraboloid" is a paraboloid... – Hawk Apr 30 '12 at 0:25
Ah. Getting used to the way the textbook authors write things will help. Even now we're being a little sloppy, because in principle "the paraboloid" is the full infinite locus and the bit we have here ought be be "the truncated paraboloid" or perhaps "the segment of a paraboloid shown". – dmckee Apr 30 '12 at 0:37

A closed surface such as this one can be thought of two open surfaces. One open surface is the circular disk. The other open surface is the paraboloid. So think of the total flux as the flux through the circular disk plus the flux through the paraboloid, remembering that the total flux must be zero.

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