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!