Determain thickness of layers in cylindrical capacitor

[Vernes]

Homework Statement

We have two coaxial cylinders with radius r₀ and r₁. The space between the two cylinders is completely coverd with two coaxial isolation layers with relative dielectric constants ε₁ and ε₂, ε₁ is for the inner layer. Calculate the thickness of the inner layer such that the voltage drop over both layers is equal.

Homework Equations

$$\oint \textbf{D} \cdot d\textbf{S}$$
$$\textbf{D} = \epsilon_0 \epsilon_r \textbf{E}$$
$$V(r_2) - V(r_1) =- \int_{r_1}^{r_2} \textbf{E} \cdot d\textbf{l}$$

The Attempt at a Solution

Let Q_inside be the free charge inside where r < r₀ and let δ be the thickness of the inner layer.
Then from symmetry we get when r₀ < r < r₁ :
$$\textbf{D(r)} = \frac{Q_{inside}}{4\pi r^2} \hat{r}$$
So the electric field E₁ for r₀ < r < (r₀ + δ) is:
$$\textbf{E}_1 = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_1 r^2} \hat{r}$$
and the electric field E₂ for (r₀ + δ)< r < r₁ is:
$$\textbf{E}_2 = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_2 r^2} \hat{r}$$
The voltage drop over the first layer is given by:
$$V(r_0 + \delta) - V(r_0) =- \int_{r_1}^{r_1 + \delta} \textbf{E}_1 \cdot d\textbf{l} = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_1}(\frac{1}{r_0 + \delta} - \frac{1}{r_0} )$$
The voltage drop over the second layer is given by:
$$V(r_1) - V(r_0 + \delta) =- \int_{r_1 + \delta}^{r_2} \textbf{E}_2 \cdot d\textbf{l} = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_2}(\frac{1}{r_1} - \frac{1}{r_0 + \delta} )$$
The voltage drops should be equal to we get:
$$\frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_2}(\frac{1}{r_1} - \frac{1}{r_0 + \delta}) = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_1}(\frac{1}{r_0 + \delta} - \frac{1}{r_0}) \Leftrightarrow \delta = \frac{(r_1-r_0)\epsilon_1 r_0} {\epsilon_2 r_1 + \epsilon_1 r_0}$$
This is not the right answer according to the key but i can't seem to find where I go wrong.

 

[Vernes]

Ok, I see one error that i did. D should be:
$$\textbf{D} = \frac{Q_{inside}}{2\pi L r}\hat{r}$$
Similar steps as before gives me:
$$\frac{1}{\epsilon_1} \ln \frac{r_0+\delta}{r_0} = \frac{1}{\epsilon_2} \ln \frac{r_1}{r_0+\delta}$$
My remaining question is then how do i solve for δ in the equation above?

Edit: Ok, I was a bit to hasty, it wasn't so hard to solve for δ as i expected...
If anyone is interested I got:
$$\delta = {r_0}^{ \frac{\epsilon_2}{\epsilon_1 + \epsilon_2}} \cdot {r_1}^{ \frac{\epsilon_2}{\epsilon_1 + \epsilon_2}} - r_0$$
Which is the right answer!