# Determain thickness of layers in cylindrical capacitor

1. Apr 19, 2015

### Vernes

1. The problem statement, all variables and given/known data
We have two coaxial cylinders with radius r0 and r1. The space between the two cylinders is completly 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.

2. Relevant equations
$$\oint \textbf{D} \cdot d\textbf{S}$$
$$\textbf{D} = \epsilon_0 \epsilon_r \textbf{E}$$
$$V(r_2) - V(r_1) =- \int_{r_1}^{r_2} \textbf{E} \cdot d\textbf{l}$$
3. The attempt at a solution
Let Qinside 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.

2. Apr 19, 2015

### 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!

Last edited: Apr 19, 2015
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