<p>omg yeah!!! I just realized that we could've used the graphs for the RC circuit problem...I solved the differential equation for the circuit based on the Loop Theorem....did I do more work than I needed to??? (EM #1) Will I still get full credit if I solved it correctly?</p>
<p>What were answers to the Gauss's Law problem??? I think I did worst on that...</p>
<p>Electric field in the following regions:
r < a
a < r < 2a
2a < r < 3a
r > 3a</p>
<p>Electric Potential external to shell</p>
<p>Potential Difference (X - Y)</p>
<p>i think i got 3/4 on those, just said the electric potential was at the maximum at the surface of the shell, and didnt finish the potential difference problem...got up to where you have to plug in the distances given, so I should get 3/5 points for that.</p>
<p>That was kQ((1/y)-(1/x)), although I switched the signs and therefore lost a point.</p>
<p>it was hard...but it could have been much worse...</p>
<p>Don't think this is right but I got</p>
<p>r < a
kqr^3/R^2 <- something like that, variables might be switched around</p>
<p>a < r < 2a
kq/r^2</p>
<p>2a < r < 3a
kq/r^2</p>
<p>r > 3a
0</p>
<p>Someone confirm please?</p>
<p>Your "a < r < 2a" and "r > 3a" are definitely right, if I remember correctly. I do not recall exactly what I put down for the other two, but they were pretty complex answers. You had to use volume ratios because the problem stated that the +Q and -Q charges were equally distributed through their respective spheres.</p>
<p>well, what's the curve gonna be for this test (E and M)? near 60 for a 5? or lower?</p>
<p>Edit: For 1 (e) you did not have to draw any asymptotes or label any points because they did not say to do anything besides sketching the curve.</p>
<p>usually u do have to label the asymptote that it reaches...its only a point tho...</p>
<p>what is the Vx - Vy???</p>
<p>I got </p>
<p>Vx = kQ/a
Vy = kQ/(2a)</p>
<p>Therefore, it equals Vx - Vy = kQ/(2a)</p>
<p>And I thought the electrical potential external to the sphere would be 0 since the net charge is 0 throughout the sphere...no idea tho!!!</p>
<p>well, I find that whenever the electric field is zero the potential is at the maximum...I only said "maximum" without an explanation (ran outta time) so I dont know if I will get a point for that, probably not though...so I'm backing off even my predictions for a 30 on the FR (after messing up parts 1 b, 1 f, 3 b, and 3 d) and taking it down to a 25... 7/8 + 10 + 8</p>
<p>oooh....what did put for #2c on EM FR??? the Vx - Vy question?</p>
<p>well....I got to the substitution but left them as Y and X instead of 2a and a so I probably lost 2 outta 5 points for that.</p>
<p>Still...anyone know the e and m curve? My physics teacher says it's in the upper 50s for a 5....</p>
<p>My teacher did all the free response questions so I have all the solutions for Mechanics if any one wants to know if they're right.</p>
<p>The potential at the edge in #2 is zero because of the definition of potential:</p>
<p>Electric potential is the work per charge to bring a point charge from infinity (where V=0) to that point. Since there is no electric field from the edge to infinity, no force is acting on that point charge, and therefore no work (positive or negative) is done in moving it.</p>
<p>really? I dont know why but i remember in example where the center of a square of four equal charges (of equal sign too) had a 0 E field but a quantified potential....</p>
<p>piccolojunior...the Gauss sphere had a net charge of 0....the design you are talking about has a net charge of like +4....different situation</p>
<p>dont worry about it tho...its only like a point or two off...sounds like ur still gonna hang onto ur 5!!! (thats awesome by the way...hopefullly ill be the same :-)</p>
<p>Also, piccolojunior, while superposition of the fields created by the charges in your example would result in a net field of zero because E is a vector, the voltage would be a simple scalar sum of the potentials from each charge.</p>
<p>e.g.:
field: 1+1-1-1
voltage: 1+1+1+1</p>
<p>I hope so, but then again since the MC was easy I shouldnt really be panicking. :)</p>
<p>edit: damn, that's right, I just scribbled down "max" because I had no time left but common sense should have told me to put 0 because there was no implied calculation...oh well, what's done is done and hopefully I stil have enough for a 5 (like say a 63 or so)</p>
<p>july!!!!!!!!!!!!!</p>
<p>yes...admanrich can u please post your "official" answers from ur teacher???</p>