<p>Hmm ok what you said for part b makes sense now that I think about it, but how does ∫E~dA equal 0 if E doesn’t equal 0? Because we’re agreed that there’s no charge enclosed…so even if you can’t bring the E out something still needs to equal 0? (unless maybe the field at one point negates the field at the other for the entire surface?)</p>
<p>And why isn’t part d’s answer A? The question said it was a conducting sphere with charge +Q, so the charge would distribute itself around the edge of the sphere leaving the center uncharged…</p>
<p>And part f I don’t think I did right because if we only pay attention to the 1/8th that the problem mentions, then technically there would be flux on all 6 sides, so the proportion I set up would be wrong…</p>
<p>(ps thanks for your help and for responding, Equuleus and NYEM)</p>
<p>Ah, the part b problem was seriously flawed because in reality, there is no such thing as non-uniform charge distribution. Charges always distribute themselves uniformly, almost instantly. The question was hypothetical, and more a test of vector calculus than your actual understanding of the physics of charge.</p>
<p>The integral E dot dA does equal zero but that doesn’t mean E needs to be zero. E can be coming out at some points and then going out at other points so that the total flux is zero. If it were zero everywhere, that would essentially imply a uniform charge distribution which is against the hypothesis of the question.</p>
<p>Part d was another horrible question and I personally believe its ambiguity will ellicit some contradicting answers yet answers that have their cogency. I’m probably wrong on it though.</p>
<p>For the question about normal/centripetal force, the gravity of the body has a component that is perpendicular to the direction of motion/parallel to the normal force and thus contributes to the centripetal force. I don’t remember exactly what the answer ended up being, but you had to take that into account.</p>
<p>Anyway, my answers. Some are definitely wrong:</p>
<p>E&M. 1.</p>
<p>(a) Surface is sphere with r<R and goes inside shell. Basically, the surface encloses no charge whatsoever. By Gauss’ Law, if no charge is enclosed, electric field must be 0.
(b) Yes. While the charge density on the outside of the sphere is not uniform, the Gaussian surface still resides within the shell, in which no charge is enclosed, so E is still 0.
(c) I said none, since I reasoned that no matter what, it would be impossible for all field lines to be parallel to the face. But I agree that it’s all faces with A
(d) Corner F
(e) E=(kQ)/(3L^2)
(f) I did some really random stuff, integrating E dA. I said A=L^2, so dA=2LdL. I subbed that in, said that E=(kQ/8L^2) and I integrated from L to L*sqrt(3). Definitely got something weird. Hopefully I get points.</p>
<p>E&M. 2.</p>
<p>(a) i. Q=CV. Make sure units of capacitance are farads and not mF.
ii. Q=0 from t=0 to t=t<em>1. Increases exponentially (concave down) with a horizontal asymptote of CV (or equivalent number)
iii. I=0 from t=0 to t=t</em>1. I jumps up to V/R (or equivalent number) at t=t_1 and decreases exponentially (concave up) with horz. asy. of I=0.
(b) i. U=.5QV. Modify so that you have Q and C and solve (I may have plugged in 9 for V without thinking…ugh)
ii. Conservation of energy. The energy from part (b)i equals .5LI^2, with L being the inductance. Solve for I (I got .297 A).
iii. Use V=Q/C to solve for voltage. V=-L(dI/dt). Solve for dI/dt.</p>
<p>E&M. 3.</p>
<p>(permittivity constant/2pi)=(2E-7)</p>
<p>(a) i. B=0, since dl is zero since no current flows in that section.
ii. B=(2E-7)(I)(r^2/b^2)
iii. B=(2E-7)(I/4b)
(b) I put that it went horizontally to the right, though it should be pointing right and down.
(c) No force. No electric field is present, so no electric force. For magnetic force to be felt, electron must be moving (F=qvB). However, v=0, so magnetic force=0.
(d) i. Graph
ii. Plug in pi, b, and I values from directions. Get B/r from the slope of line. Solve for permittivity constant (I got something like 1.46E-7; it was very close to the real value).</p>
<p>I think I agree with most of your answers, 314, except that I had no idea how to do LC circuits. One thing I had different was 3.a.ii. I had (2E-7)(I)((r^2-a^2)/(b^2-a^2)), I believe. And I got a different permeability constant estimation, but that’s likely because we drew slightly different lines of best fit.</p>
<p>This is an amazing article, incredibly in depth on Gauss’ Law. Although sadly it doesn’t mention shells of non-uniform charge. We’ll just have to wait 'til the collegeboard puts up the answers. Any idea when that might be dude?</p>
<p>OH part f was like this:
The charge enclosed is Q/8 so the total flux is Q/8 divided by epsilon zero.
Since each square has the same area, and there are 6 squares, each square has flux equal to the total flux, divided by 6, which gives Q/48 divided by epsilon zero.</p>
<p>@NYEM: I think the part f is asking a completely different thing from part e because in part f, the Q/8 is enclosed within the cube. If it is enclosed in the actual inside of the surface then it does contribute flux to all faces of the surfaces and since all those faces have the same area, the field coming out of them must be the same.</p>
<p>^I don’t think that was the case, Equuleus. From the test:
The eighth of the sphere is still at corner A - If the entire sphere doesn’t contribute flux to three surfaces indicated it part (c), this eighth of the sphere also will not. The question doesn’t indicate that the eighth of the sphere has been moved into the center of the cube - in which case you would be right. </p>
<p>For the AP Physics C: Mechanics form B (which is the late testing version), on question 2, I accidentally made the normal force = mgsin(theta) instead of mgcos(theta), which made my horizontal forces mgcos(theta). I consistently did this throughout the question. Will I receive no credit for deriving an expression of the net force in the horizontal direction because I put the wrong trig function? Please someone answer me, I’m dying to know</p>
<p>^Form B is international version. Form A is late-testing. And if you used an incorrect answer correctly in a subsequent part, you will receive full points in the subsequent part. You are penalized only once per mistake.</p>