Let the length of the cylinder be $h$. To find the total charge on the core of radius $a$, first find the charge on a cylindrical shell of infinitesimal radius, $dr$:
$dq=\rho *2 \pi h r dr=-k*2 \pi h dr$
$q=\int_{0}^{a}{-k*2 \pi h dr}=-2 \pi hak$
Total charge on the outer cylinder $=\sigma * 2\pi bh$
Equate the two,
$\sigma * 2\pi bh=2 \pi hak$
$\sigma=\frac{ak}{b}$
You were mostly on the right path vis-à-vis solving b)
However, note that you are dealing with volume charge.
Using the same reasoning as above, the total charge inside the cylinder with $r<a = -2 \pi hrk$
Using Gauss' law, $\phi_e=\frac{q}{\epsilon_0}$
$\Rightarrow E*2\pi rh=-\frac{2 \pi hrk}{\epsilon_0}$
$\Rightarrow E=-\frac{k}{\epsilon_0}$
$\vec {E}= <\frac {-k}{\epsilon_0} \frac {x}{(x^2+y^2)^{\frac {1}{2}}}, \frac {-k}{\epsilon_0} \frac {y}{(x^2+y^2)^{\frac {1}{2}}},0>$
@ArtforLife 's statement:
for electrostatic case, the field for r < a must be zero if the inner core is a conductor. If it isn't, you can calculate using Gauss's law.
is redundant. If the cylinder had been a conductor, there wouldn't have been a volume charge density in the first place. All the charge would be on the surface.
When $a<r<b$, net $q$=$-2 \pi hak$
$\phi_e=\frac{q}{\epsilon_0}$
$\Rightarrow E*2\pi rh=-\frac{2 \pi hak}{\epsilon_0}$
$\Rightarrow E=-\frac{ak}{\epsilon_0 r}$
$\vec {E}= <\frac {-ak}{\epsilon_0} \frac {x}{(x^2+y^2)}, \frac {-ak}{\epsilon_0} \frac {y}{(x^2+y^2)},0>$
Best Answer
The problem is easier than you think, and you don't have to worry about the thickness because your interested in the field OUTSIDE the conductor. I'm sure by now you know that the easiest way to solve this is by using Ampere's law. Our closed loop will be a circumference of radius r (distance of point P to the center) and Ampere's law tells us that: $$\oint _C {\bf B}\cdot d{\bf l}=\mu _0 I_{enclosed}$$
As for the first integral, since B is uniform through our curve, it will simply become $$B2\pi r$$
Now for $I_{enclosed}$ what is it? It's the current the traverses our closed loop. Now if we have a current I going on one direction in one conductor, and the same current I going in the opposite direction in outer conductor the currents will cancel out (notice that they give you the total current I not the density of current (meaning that the dimensions don't matter).
So that will lead up to a B of zero. The remarkable conclusion is that it doesn't matter how thick the outer or the inner conductor are, if they are traversed by the same current but in opposite directions the B field will always be zero. I hope this helped.
Inside the cable the scenario will be different if you want to we can work on that too.
EDIT1: OOPS I forgot to add the permeability of vacuum on Ampere's law. My apologies
EDIT2: So let's see what's happening inside the cylinder. For that we take the same considerations as before: circular closed loop centered in the axis of the cable now with an arbitrary radius r.
For all of the cases we will consider, the left side of Ampere laws always results on
$$B2\pi r$$
So the difference will lie on our $I_{enclosed}$
For $r<a$ you will need the current density that goes through that current. That will be given by the total current divided by the section on the conductor (think of it as water through a pipe, the flux of water in that quantity divided by the section of the pipe). That is $\frac{I}{\pi a^2}$. Now that you have your density you multiply it by the section of your current loop (you now think of the loop as a disk). You will then have $$I_{enclosed}=\frac{I \pi r^2}{\pi a^2}$$
Putting everything together ´
$$B2\pi r=\frac{\mu_0 I \pi r^2}{\pi a^2}$$
so
$$B=\frac{\mu_0 I r}{2 \pi a^2}$$
The direction of B is given by the right hand rule, so it's clockwise.
Now the situation of a
For b
And putting everything together leaves you with $$B=\frac{\mu_o I}{2 \pi r}(1-\frac{r^2-b^2}{c^2 - b^2})$$
The direction of B I will also leave up to you now :)
Tell me if there were any questions left, did you understand everything?