<p>Well, since nobody else is stepping up, I guess I’ll kick off the E&M attempts.</p>
<p>(I’m using ~ as dot product…because I can’t figure out how to type it otherwise…)</p>
<p>1.</p>
<p>a. ∫E~dA=Qenc/epsilon, so pick any/every spherical surface within the shell and integrate, finding that Qenc is always zero, so E is also always 0 (0 * 2piR = 0).</p>
<p>b. Yes, the field is still zero everywhere in the sphere. Basically just repeat ∫E~dA=Qenc/epsilon for the exact same Gaussian surface(s) and you find that Qenc is still zero, so E is still zero.</p>
<p>c. Flux = 0 for ABCD, ABGH, ADEH. Flux = ∫E~dA, so when the angle between E and dA is 90 degrees, the dot product is 0 (cos90=0). This works because the sphere can be treated as a point charge located directly at point A.</p>
<p>d/e. The field has the least magnitude at corner A, because again, ∫E~dA=Qenc/epsilon (Gauss’ Law), so pick an infinitely small spherical surface that only surrounds the point at corner A. All of the sphere’s charge will be outside of this surface because charge moves to the outside of a conducting object. Hence, Qenc = 0 so E = 0 as well.</p>
<p>f. Ummm I don’t know/think I got this right, but I set ∫E~dA=Q/8epsilon, because only one eighth of the charge was within the cube. Then I said that each of the 3 sides with flux would receive the same flux, so ∫E~dA for side CDEF would just be 1/3 of the total flux from that portion of charge, so I divided Q/8epsilon by 3 to get my answer of Q/24epsilon. Again, I doubt what I did here was right, but maybe someone else can tell me how it’s supposed to be done.</p>
<p>Phew, anyone else care to post #2 or #3? Please? :D</p>