Coaxial cable conductors with dielectric: polarization of charge

[Granger]

Homework Statement

Consider a coaxial cable which consists of an inner cylindrical conductor of radius R1, and a shell cylindrical conductor of radii R2 and R3. The 2 conductors are separated with a dielectric material of permittivity ε. Consider the length of the cable, ℓ, much larger than R3. The cable is connected to a voltage source V. Determine the polarization charge distribution per unit of length.

Homework Equations

3. The Attempt at a Solution [/B]

So I have no problem calculating the surface polarization charge distribution.
Having:

$$D=\frac{\lambda}{2 \pi r}$$
$$E=\frac{\lambda}{2 \pi \epsilon r}$$
$$P=\frac{\lambda (\epsilon - \epsilon_0)}{2 \pi r \epsilon}$$
$$V=\frac{\lambda}{2 \pi \epsilon} \times \log{\frac{R_2}{R_1}}$$

Than we should have in r=R1:

$$\sigma= -\frac{V (\epsilon - \epsilon_0)}{R_1 \log{\frac{R_2}{R_1}} }$$

Now they as for the density by unit of length and the answer should be:

$$\lambda= -\frac{2 \pi V (\epsilon - \epsilon_0)}{\log{\frac{R_2}{R_1}} }$$

According to the answer, the relation between lambda and sigma should be:

$$\lambda = 2 \pi R_1 \sigma$$

However I don't understand how it can be?
So my question is exactly how to get to that relation between lambda and sigma (if that relation is correct, of course).
Thanks!

 

[kuruman]

This equation $\lambda = 2 \pi R_1 \sigma$ is dimensionally incorrect. Probably a typo. $R_1$ must be in the denominator. Then it will be consistent with what you got.

On edit: The relation is not dimensionally incorrect. Sorry, I got confused.

 

[Granger]

This equation $\lambda = 2 \pi R_1 \sigma$ is dimensionally incorrect. Probably a typo. $R_1$ must be in the denominator. Then it will be consistent with what you got.

But if R1 is in the denominator then I will not get to the answer.
Do you know how should I proceed?

 

[kuruman]

You have an expression for $V$ and an expression for $\sigma$. Did you not derive these? What happens if you put your expression for $V$ in your expression for $\sigma$?

 

[Granger]

Yes what I did was I took my expression for $V$, isolated $\lambda$ in that expression and substituted in my expression for $P$. That was how I got to my expression for sigma (with a minus sign).
If I substitute V back to my sigma expression I will have:

$$\sigma= -\frac{\lambda (\epsilon - \epsilon_0)}{R_1 2 \pi \epsilon}$$

And equaling this new expression to the expression I had for $\lambda$ before:

$$\lambda= -\frac{2 \pi \epsilon}{\log {R1/R2}}$$

Which is different from the expression for lambda I got before!
What am I doing wrong?

 

[haruspex]

You did not quote any relevant equations, but appear to have used quite a few.
What equation did you use to obtain V from P?

 

[kuruman]

Actually, you can get the polarization charge density using $\sigma=\vec{P} \cdot \hat{n}_{out}$ without figuring out the voltage. Unfortunately the result is the same as in post #5. However, if you use the boundary condition $(\vec{D}_2-\vec{D}_1)\cdot \hat{n}=\sigma_{free}$ and that $\vec{D}_1=0$ at the conductor, you get $\frac{\lambda}{2 \pi R_1} =\sigma_{free}$. It appears then that the $\sigma$ mentioned in the relation given by the solution is $\sigma_{free}$.

 

[haruspex]

Actually, you can get the polarization charge density using $\sigma=\vec{P} \cdot \hat{n}_{out}$ without figuring out the voltage.

I don't understand. The given voltage is the only way to find $\vec{P}$, no?

 

[kuruman]

I don't understand. The given voltage is the only way to find $\vec{P}$, no?

In post #1, OP found D using Gauss's Law, where λ is the linear free charge density. Then apparently OP combined $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$ and $\vec{D}=\epsilon \vec{E}$ to find $P=\frac{\lambda (\epsilon - \epsilon_0)}{2 \pi r \epsilon}$. Is there a pitfall with OP's reasoning? I don't see one, but I have fallen into pits before.

 

[haruspex]

In post #1, OP found D using Gauss's Law, where λ is the linear free charge density. Then apparently OP combined $\vec{D}=\epsilon_0 \vec{E}+\vec{P}$ and $\vec{D}=\epsilon \vec{E}$ to find $P=\frac{\lambda (\epsilon - \epsilon_0)}{2 \pi r \epsilon}$. Is there a pitfall with OP's reasoning? I don't see one, but I have fallen into pits before.

But λ is to be found. The givens are V, the radii and the permittivity. The OP found a relationship between V and λ as a way of finding λ, not as a way of finding V.

 

[kuruman]

Yes, in the equation $P=\frac{\lambda (\epsilon - \epsilon_0)}{2 \pi r \epsilon}$, $\lambda$ should be more correctly written as $\lambda_{free}$. It is not the polarization linear charge density. The voltage is needed to eliminate $\lambda_{free}$ from the equation.