Electric field in a coaxial cable using Gauss's law

[jimithing]

A long coaxial cable consists of an inner cylindrical conductor with radius $$a$$ and an outer cylindrical shell of inner radius $$b$$ and outer radius $$c$$. The cylindrical shell is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length $$\lambda$$.

Calculate $$E$$
a) at any point between the cylinders
b) at any point outside the cylindrical shell
c) Find the charge per unit length on the inner surface and outer surface of the shell.

a) wasn't a problem, found $$E = \frac{\lambda}{2\pi rh}$$
b) and c) I'm having problems with

would $$E$$ outside the shell be 0 since the insulating shell carries no charge?

 

[Claude Bile]

You answer to a is wrong for one simple reason (among others), it is dependent on the Gaussian Surface you chose, when it shouldn't be. The total charge, q is not lambda, but lambda times the height of your Gaussian surface (think about the units, lambda is in C/m, you need to multiply by a distance to get C). Also there should be an epsilon 0 in there if you are working in S.I. units.

Claude.

 

[M98Ranger]

For part b), you are correct that the electric field outside the cylindrical shell would be 0. This is because Gauss' law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. Since there is no net charge within the cylindrical shell, the electric flux through any closed surface outside the shell would be 0, resulting in an electric field of 0.

For part c), we can use Gauss' law again to find the charge per unit length on the inner and outer surfaces of the cylindrical shell. We can imagine a Gaussian surface in the shape of a cylinder with radius r, centered between the inner and outer surfaces of the cylindrical shell. The enclosed charge within this surface would be the charge per unit length on the inner and outer surfaces multiplied by the length of the cylinder, which is 2\pi rh.

Using Gauss' law, we can equate the electric flux through this surface to the enclosed charge divided by the permittivity of free space. This would give us the following equations:

For the inner surface:
\frac{Q_{inner}}{\epsilon_0} = \frac{\lambda_{inner} \cdot 2\pi rh}{\epsilon_0} = \oint \vec{E} \cdot d\vec{A} = E \cdot 2\pi rh

Solving for \lambda_{inner}, we get:
\lambda_{inner} = \frac{E}{2\pi r}

Similarly, for the outer surface:
\frac{Q_{outer}}{\epsilon_0} = \frac{\lambda_{outer} \cdot 2\pi rh}{\epsilon_0} = \oint \vec{E} \cdot d\vec{A} = E \cdot 2\pi rh

Solving for \lambda_{outer}, we get:
\lambda_{outer} = \frac{E}{2\pi r}

Therefore, the charge per unit length on both the inner and outer surfaces of the cylindrical shell would be equal to the electric field at that point divided by 2\pi r. Since the electric field outside the shell is 0, the charge per unit length on the outer surface would also be 0. On the other hand, the charge per unit length on the inner surface would be equal to the electric field at that point divided by 2\pi r.

I hope this helps clarify