Q:
In the beautiful
figure shown, a nonconducting spherical shell with inner
radius a and outer radius b has a volume charge
density of 
= A/r, where A is a
constant and r is the distance from the center of the
shell. In addition (just to make things interesting) a point
charge of qo is located at the center of the
sphere.
Derive the equation which would give the magnitude of the E-field
in the three regions shown.
| Non-conducting means that
the charge stays in place. A conducting shell would have had charges which were free to move within the shell, following their natural electrical urges - but that's a topic for a different problem. |
You gotta just love spherical symmetry, folks. Any time you've got a sphere, Gauss' Law comes to the rescue.
The integral means "Add up everything inside me". The "me" part depends on how far away you want to be from the center of the sphere. This arbitrary distance is what we called r. E stands for the electric field at a distance of r from the center. dA stands for a piece of area of the spherical Gaussian surface that you're supposed to imagine exists with the very same radius r.
Inside the integral is the flux (E · dA) which involves the dreaded dot product. For those of you who wanted to build bridges and clean sewers for a living, vector calculus can seem cold-hearted and cruel. Fortunately, for spherical surfaces, the E vector always points in the same direction as the dA vector. This means that the dot has no power over you:
This side of Gauss' Equation (the left side) is the same for all three regions shown. The only thing that we'll have to worry about is the right hand side of Gauss' Law which deals with the amount of charge that is enclosed within the Gaussian surface.
Region I - Inside the shell
The only charge enclosed within a Gaussian surface of radius r
in this region is that point charge, qo. To
find the equation for the E - field is rather easy:
As long as you stay inside the shell, you can use
this equation to determine how strong the E-field
will be.
Region
II - Within the shell
This is the tough part of the question. You see, we're given the charge
density (how much charge can be found in a given
volume) which varies with how far you are from the center. With,
= A/r, the further away you are, the less charge
there is. Well, as ugly as it seems, you're gonna have to add up
the charge that's within the shell. Since the charge density
changes with your position in the shell, adding it up is a job
for an integral.
The beginning of the whole process starts
innocently enough: 
Now, we have to figure out exactly how much charge
exists inside the Gaussian surface with radius r. Of
course you have q0. But you also have to
include the charge that exists between a and r. 
If you need a
piece of the above math explained in feel free to click here 
.
We can now finish this off: 
Okay, I'm sure you can pretty up the
final result, but this works just fine for me.
Region III - Outside of the shell
This can be done with the same procedure as above, but this time,
regardless of how far out we are, all of the charge will
be within the Gaussian surface. What will change is how we
calculate Qenc. The integral will go from a
to b, and not all the way out to r. You can only
add up charge where there's charge, and there's no charge past b.
And so,
we finally come to the end of this question: 
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