Boundary Conditions - Cylinder in dielectric

[unscientific]

Homework Statement

Part (a): List the boundary conditions
Part (b): Show the relation for potential is:
Part (c): Find Potential everywhere.
Part (d): With a surface charge, where does the Electric field disappear?

Homework Equations

The Attempt at a Solution

Part (a)

Boundary conditions are:
$$E_{in}^{||} = E_{out}^{||}$$

$$D_{out}^{\perp} - D_{in}^{\perp} = \frac{\sigma}{\epsilon_0}$$

Part (c)

First we find potential inside and outside.

1. For potential inside $V_{in}$, it is finite when r = 0, so $b_n = 0$. When r = 0, $V_{in} = 0$, so $a_0 = b_0 = 0$.

2. For potential outside $V_{out}$, as r → ∞, $V \propto \frac{1}{r}$ (From part (b)). So only term n = 1 exists.

3. For n=1, and $\phi = \frac{\pi}{2}$, $V_{in} = V_{out} = 0$, so sine term goes away: $c_n = 0$.

Now we have the two potentials:

$$V_{in} = a_1 r cos(\phi)$$
$$V_{out} = \frac{d_1}{r} cos(\phi)$$

Applying first boundary condition at r = a,

$$\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}$$

$$a_1 a^2 = d_1$$

Applying second boundary condition at r = a,

$$D_{out}^{\perp} = D_{in}^{\perp}$$
$$\epsilon_0 \frac{\partial V_{in}}{\partial r} = \frac{\partial V_{out}}{\partial r}$$
$$\epsilon_0 a_1 = -\frac{\epsilon d_1}{a^2}$$

Solving them, this implies $\epsilon_0 + \epsilon = 0$, which doesn't make sense.

 

[unscientific]

I've got a feeling that my potential inside is missing a term like the one in part (b): $V = \frac{\lambda}{2\pi \epsilon_0}\frac{d cos \phi}{r}$

Because it shouldn't strictly depend only on $cos \phi$?

 

[unscientific]

I think the potential inside should be something like:

$$V_{in} = Ar cos \phi + \frac{B}{r^2}cos \phi$$

That should solve it.