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All the above answers are correct,although none gives you an answer as to WHY you should NOT use a sphere and none addresses your "infinite gaussian surface" problem and you seem to be a bit confused on how things actually work(you have to understand how things work before you delve into the mathematics part). So,if you use a sphere,then your integral of ...

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Since the sphere has no electrical properties, I am taking it to just be a mathematical surface. In this case, we are just computing a special case of the electric flux through a disk of radius $R$ from a charge centered on its axis a distance $\ell$ away which is  \Phi = \int_0^R E(r) \cos\theta\,dS = \int_0^R dr \frac{Q}{\ell^2 + r^2} (2\pi ...

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The solid angle approach works, it's just when using that formula you need the full charge. The distance from the charge to the edge of the disc formed by your hemisphere is just $R\sqrt{2}$. So draw a sphere of radius $R\sqrt{2}$ around your point charge. The total flux through this sphere will be the full $\frac{Q}{\epsilon_0}$. Because the electric field ...

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Example 7 in this site is exactly the same as your problem http://physicspages.com/2011/10/04/gausss-law-examples/ I would like to clarify something here. Magnitude of a field of an infinitesimally-thin infinite plane is $\frac{\sigma}{2 \epsilon_0}$. We use $\sigma$ here not $\rho$ because $\sigma$ indicate the planar charge density since the plane is very ...

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Since in this problem you can exploit a symmetry (invariance by translation along the directions spanned by the slab) it is easier to compute the electric field by a wise use of Gauss' theorem. As your book says, this is quite a standard technique, which is used e.g. for the infinite plane, infinite cylinders (coaxial cables), etc.... The idea is that, as a ...

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You can use the superposition principle, in general this is more difficult as that requires you to know the charge distribution on all the objects. Note that this is not always as simple as it may seem. E.g. you say "uncharged due to attraction between two plates", but this is not the correct reason, attraction alone would not yield this, it's the inverse ...

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Gauss law is sometimes used to find the electric field, but this is usually confined to situations where there is a strong symmetry that allows you to conclude that the electric field you're looking for is precisely the one in Gauss law, and moreover it reduces to something easy to evaluate, like the product of the magnitude of $\mathbf E$ with a surface. ...

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