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Vernes

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## Homework Statement

We have two coaxial cylinders with radius r

_{0}and r

_{1}. The space between the two cylinders is completely coverd with two coaxial isolation layers with relative dielectric constants ε

_{1}and ε

_{2}, ε

_{1}is for the inner layer. Calculate the thickness of the inner layer such that the voltage drop over both layers is equal.

## Homework Equations

[tex] \oint \textbf{D} \cdot d\textbf{S} [/tex]

[tex] \textbf{D} = \epsilon_0 \epsilon_r \textbf{E}[/tex]

[tex] V(r_2) - V(r_1) =- \int_{r_1}^{r_2} \textbf{E} \cdot d\textbf{l} [/tex]

## The Attempt at a Solution

Let Q

_{inside}be the free charge inside where r < r

_{0}and let δ be the thickness of the inner layer.

Then from symmetry we get when r

_{0}< r < r

_{1}:

[tex] \textbf{D(r)} = \frac{Q_{inside}}{4\pi r^2} \hat{r} [/tex]

So the electric field

**E**

_{1}for r

_{0}< r < (r

_{0}+ δ) is:

[tex] \textbf{E}_1 = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_1 r^2} \hat{r}[/tex]

and the electric field

**E**

_{2}for (r

_{0}+ δ)< r < r

_{1}is:

[tex] \textbf{E}_2 = \frac{Q_{inside}}{4\pi \epsilon_0 \epsilon_2 r^2} \hat{r}[/tex]

The voltage drop over the first layer is given by:

[tex] 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} ) [/tex]

The voltage drop over the second layer is given by:

[tex] 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} ) [/tex]

The voltage drops should be equal to we get:

[tex] \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} [/tex]

This is not the right answer according to the key but i can't seem to find where I go wrong.