# Homework Help: Bound charges and electric displacement.

1. Nov 24, 2013

### -Dragoon-

1. The problem statement, all variables and given/known data
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.

3. 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.

2. Nov 24, 2013

### TSny

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

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

3. Nov 25, 2013

### -Dragoon-

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

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

4. Nov 25, 2013

### TSny

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

5. Nov 25, 2013

### -Dragoon-

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