Bound charges and electric displacement.

[-Dragoon-]

Homework Statement

A certain coaxial cable consists of copper wire, radius a, surrounded by a concentric copper wire of outer radius b. The space between is filled with a dielectric with a relative permittivity of $$\epsilon_{r} = \frac{s}{a}, a \leq s \leq b$$ Find the bound charges by using two different methods.

The Attempt at a Solution

First I'll find the electric displacement, the field, and the potential:

$$\oint D\cdot da = Q_{f_{enc.}} => D(2\pi sl) = Q =>D = \frac{Q}{2\pi sl}$$

Then I find the field from the displacement:

$$E = \frac{D}{\epsilon} = \frac{D}{\epsilon_{0} \epsilon_{r}} = \frac{a}{s^{2}}\frac{Q}{2\pi \epsilon_0 l}$$

Then, finally, the potential:


$$-\int_s^a E\cdot dI = \frac{Qa}{2\pi \epsilon_0 l}\int_a^s \frac{ds}{s^{2}} = \frac{Qa}{2\pi \epsilon_{0} l}(\frac{1}{a} - \frac{1}{s})$$

Now for the bound charges:

$$P = D - \epsilon_{0}E = \frac{Q}{2\pi sl}\hat{r} - \frac{a}{s^{2}}\frac{Q}{2\pi l}\hat{r} = \frac{Q}{2\pi l} = \frac{Q}{2\pi l}(\frac{1}{s} - \frac{a}{s^{2}})\hat{r}$$

The surface bound charge is:
$$\sigma_{b} = P\cdot\hat{n} = \frac{Q}{2\pi l}(\frac{1}{s} - \frac{a}{s^{2}})$$

Now, when I calculate it using a different method which involves the potential:

$$\sigma_{b} = -\epsilon_{0}( \frac{\partial V_{out}}{\partial s} - \frac{\partial V_{in}}{\partial s}) = -\epsilon_{0}(\frac{Q}{2\pi \epsilon_{0} l}\frac{a}{s^{2}} - 0) = -\frac{Q}{2\pi l}\frac{a}{s^{2}}$$

Which gives one of the terms correctly, but the other term is clearly missing from here. Any hints on where I'm going wrong in this? Thanks in advance.

 

[TSny]

The surface bound charge is:
$$\sigma_{b} = P\cdot\hat{n} = \frac{Q}{2\pi l}(\frac{1}{s} - \frac{a}{s^{2}})$$

Shouldn't the bound surface charge only be defined for s = a or s = b?

Also, there will be a bound volume charge density ρ_b for a < s < b.

 

[-Dragoon-]

Shouldn't the bound surface charge only be defined for s = a or s = b?

Also, there will be a bound volume charge density ρ_b for a < s < b.

Interesting, so then for s = a, the bound charges are zero? And for s = b, we have:

$$\sigma_{b} = - \frac{Q}{2\pi l}(\frac{1}{b} - \frac{a}{b^{2}})$$

 

[TSny]

Yes, that looks right to me, except I get the opposite sign. I guess the sign depends on which conductor carries positive Q.

 

[-Dragoon-]

Yes, that looks right to me, except I get the opposite sign. I guess the sign depends on which conductor carries positive Q.

Alright I see, thanks for the help once again, TSny.