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There is an infinite slab of charge, and a (Gaussian surface) cylinder whose ends are both outside of the slab.

$\phi_A$ is the flux through this cylinder, by symmetry the component of the flux through the top end is $\frac{ \phi_A}{2}$.

Now, lift bottom end of the cylinder until it is inside the slab. The net flux is now $ \phi_B$.

Proposition 1: The component of the flux through the top end is unaffected by the position of the bottom end (even though the total flux is), as the electric field and area at and of the top end are both unaffected by the movement of the bottom end.Thus the flux through the top end is still $\frac{ \phi_A}{2}$.

Proposition 2: Therefore the flux through the bottom part is all the flux that is remaining (that is, $\phi_B-\frac{ \phi_A}{2}$).

This must be balderdash, as with by extension of my logic, the flux through the bottom in the second case is$\phi_B-(\phi_B-\frac{ \phi_A}{2})=\frac{ \phi_A}{2}$: that is, the net flux is unaffected as the top and bottom flux add up to $ \phi_A $, in contradiction with $\phi=\frac{Q}{\epsilon_0}$. What is incorrect with propostion 1, both physically and mathematically?

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It's actually better if you keep the part that you deleted in your last edit, because without it, it doesn't make much sense why you think something is wrong with proposition 1. (You can just roll back to the previous revision) –  David Z Oct 11 '12 at 19:38
    
I meant to, but wasn't sure how. Thanks! –  Alyosha Oct 11 '12 at 19:57

1 Answer 1

up vote 1 down vote accepted

Propositions 1 and 2 are both fine. However, it doesn't follow that the flux through the bottom in the second case is $\frac{\phi_A}{2}$. When you move the bottom of the cylinder up into the slab of charge, the flux through the bottom decreases because of the decrease in the amount of charge contained in the cylinder.

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Thank you, I got confused, using the bottom's flux where I should have used the top's. –  Alyosha Oct 11 '12 at 19:36

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