I have a very long coaxial cable with a cylindrical core of radius $a$ that has negative charge density $\rho = -k/r$ where $r$ is the distance from the axis of the cylinder, and an outer shell of radius $b$ that carries a positive surface density $\sigma$, chosen so that the net charge of the coaxial cable is 0.
a) How do I express $\sigma$ in terms of $k$, $a$, $b$? I was thinking of integrating to get the current, but once I have that I don't know where to go further.
b) What is the electric field for $r<a$ and $a<r<b$? I am trying to find the linear charge density for radius s from the axis like this: $\int_{0}^{s}-\frac{k}{r}*2\pi*r dr = -sk2\pi$ And then I would use Gauss' Law. This doesn't seem right to me though.
Best Answer
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:
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 $q$=$-2 \pi hak$
$\phi_e=\frac{q}{\epsilon_0}$
$\Rightarrow E*2\pi rh=-\frac{2 \pi hak}{\epsilon_0}$
$\Rightarrow E=-\frac{ak}{\epsilon_0 r}$
$\vec {E}= <\frac {-ak}{\epsilon_0} \frac {x}{(x^2+y^2)}, \frac {-ak}{\epsilon_0} \frac {y}{(x^2+y^2)},0>$