# Electric field inside coaxial cable

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.

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 $q$=$-2 \pi hak$

$\phi_e=\frac{q}{\epsilon_0}$

$\Rightarrow E*2\pi rh=-\frac{2 \pi hak}{\epsilon_0}$

$\Rightarrow E=-\frac{ak}{\epsilon_0 r}$

$\vec {E}= <\frac {-ak}{\epsilon_0} \frac {x}{(x^2+y^2)}, \frac {-ak}{\epsilon_0} \frac {y}{(x^2+y^2)},0>$

a) consider a segment of the cable of length L. Calculate the charge contained on the inner cylindrical core of radius a. Then set the same charge to the outer cylinder of radius b. Calculate sigma from this charge and the area of cylinder with radius b and height L.

b) 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.

For the area between the cylinders, the field will only be due to the inner cylinder. Since the distribution of charge is symmetric around the axis of the inner cylinder, you can indeed transform the problem to that of a linear charge density. I think that is the idea you had. Calculate the amount of charge contained on the inner cylinder per unit length. Then, apply Gauss's law.