Concentric conducting cylinders

[hmparticle9]

Homework Statement

Consider a hollow conducting cylinder parallel to the $z$-axis, of radius $a$ and charge $\lambda$ per unit length surrounded by an outer hollow conducting cylinder of radius $b$ with charge $-\lambda$ per unit length.

(i) Find the field for all $r$
(ii) What is $\sigma$, the charge per unit area on the inner cylinder?
(iii) Consider the field between two cylinders when $(b-a) << a$ is very small and compare the field to that inside a parallel plate capacitor.

Relevant Equations

Gauss's law:
$$\int_{S} \mathbf{E} \cdot \text{d}\mathbf{S} = \frac{q}{\epsilon_0}$$

To make a start. In part (i) we can use Gauss's law to show that inside the smaller cylinder and outside the larger cylinder the field $\mathbf{E} = 0$. I am not sure how to progress in the remaining case.

This is like the last problem I posted in the sense that I find it hard to get off the mark. Could you offer me some advice?

 

[kuruman]

Why don't you post your attempt for part (i)? We'll take it from there.

 

[hmparticle9]

Aside from what I have said above I can't find anything to help me in my lecture notes. Here is where I am getting my knowledge from: https://oyc.yale.edu/physics/phys-201/lecture-4

The last 15 mins or so of lecture 4 talk about conductors

 

[kuruman]

How about using Gauss's Law to answer part (i). You said

In part (i) we can use Gauss's law to show that inside the smaller cylinder and outside the larger cylinder the field $\mathbf E=0.$

Follow a similar process except that your Gaussian surface is a coaxial cylinder that encloses the smaller cylinder but not the larger cylinder.

 

[hmparticle9]

I have solved part (i).

$$\int_S E(r)\mathbf{e}_r \cdot \mathbf{e}_r dS = E(r) \int_S dS = E(r) 2 \pi r h = \lambda \frac{h}{\epsilon_0}$$

Hence

$$E(r) = \frac{\lambda}{2 \pi r \epsilon_0}$$

 

[kuruman]

I have solved part (i).

$$\int_S E(r)\mathbf{e}_r \cdot \mathbf{e}_r dS = E(r) \int_S dS = E(r) 2 \pi r h = \lambda \frac{h}{\epsilon_0}$$

Hence

$$E(r) = \frac{\lambda}{2 \pi r \epsilon_0}$$

Very good. How much charge, do you think, is enclosed by this Gaussian surface?

 

[hmparticle9]

I am not sure. We know that the inner cylinder has charge $\lambda$ per unit length.

$$\lambda h = \sigma 2\pi a h \implies \sigma = \frac{\lambda}{2 \pi a}$$

This answers (ii)

 

[hmparticle9]

Now for part (iii). In that case we have a massive "ring". If we are local to the space between the two conductors then it looks to us that we are in between two planes. One with positive charge and the other with negative charge???

I understand now. $r \approx a$ Hence
$$E(r) \approx \frac{\lambda}{2 \pi a \epsilon_0} = \frac{\sigma}{\epsilon_0}$$