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The book says it is E0πr2 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.
Answer
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 E0πr2.
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=Fend-cap+Fparabaloid=−E0πr2+Fparabaloid
or, if we re-arrange things
Fparabaloid=E0πr2.
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