# capa4 - Phys 1120, CAPA #4 solutions. 1) Look at last...

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Phys 1120, CAPA #4 solutions. 1) Look at last week's similar problem, near the end. This is pretty much the same. Here's the big picture: Gauss' law tells us that Flux is proportional to Q(inside). In this problem, symmetry says E must be radial everywhere. Thus, if you know E is zero on some spherical surface, then you conclude Q(inside that surface) must vanish. We know E is zero in the "grey regions", because E is ALWAYS zero inside conductors in electrostatics problems. ("inside conductors" really means PHYSICALLY INSIDE the actual conductive material. The grey regions in this figure, not the "holes" or spaces between them!) Since there's ZERO charge in the central region (it's empty!), I immediately conclude E=0 up to radius a, and then we're in a metal, so in fact E=0 all the way out to radius b. That means the given "charge on the inner shell", which I will call q1, must all reside on the OUTER edge of that conductor, i.e. right there at r=b. This is normal for metals - if you put charges on them, and there's nothing inside, the charge moves to the OUTSIDE. (After all, charges repel, they WANT to get as far apart as they can, and they're free to roam through a conductor, so they go to the outside if there's nothing attracting them inwards) As we move on out, there IS now charge inside of us, so there IS an E field in the region from r=b to r=c. (Given by k q1/r^2, Coulomb's law, or Gauss law. ..). Next, we reach r=c. The E field has to disappear for the next region, so there must be some OPPOSITE charge accumulated at r=c for the E field lines to end on. In fact, using Gauss law anwhere in this second grey region (where E=0, remember!) we conclude q(enclosed)=0. Now, we know we enclose q1 sitting on the inner shell, so there must be -q1 piled up on the inner surface of the outer shell, i.e. -q1 sits at r=c. (Remember, for static situations, charge on conductors always accumulates on surfaces, either inner or outer, never in the volume. ..) E=0 from r=c to r=d ('cause we're in a conductor), and then we get to r=d. Now what? Well, remember, the outer conductor had a grand total charge of q2 given in the problem statement. If there's - q1 on the inner surface, but q2 total, you HAVE to have q2+q1 on the outer surface, so that you have a grand total of -q1 + (q2+q1) = q2. (You put q2 on that conductor, it can spread out, and some can head to the middle, but the grand total has to remain what you put on it in the first place) That's probably the conceptually hardest part of this whole problem - make sure you get it! Ask someone/talk/think more about it if you don't!

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## This note was uploaded on 10/08/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

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capa4 - Phys 1120, CAPA #4 solutions. 1) Look at last...

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