Irodov 3.107: The Inner Layer Sets the Limit – Amit & Deepali Sinha Foundation

Problem Irodov 3.107

A cylindrical capacitor has an inner electrode of radius $R_1$ and an outer electrode of inner radius $R_3$ . The gap holds two coaxial dielectric layers: permittivity $\varepsilon_1$ from $R_1$ to $R_2$ , and permittivity $\varepsilon_2$ from $R_2$ to $R_3$ . The breakdown fields of the two dielectrics are $E_1$ and $E_2$ respectively. Find the breakdown voltage of the capacitor, given that $\varepsilon_1 R_1 E_1 < \varepsilon_2 R_2 E_2$ .

The double-layer cylindrical capacitor in cross-section about its axis O: the inner electrode of radius R_1, the layer ε_1 out to R_2, the layer ε_2 out to the outer electrode at R_3. — The double-layer cylindrical capacitor in cross-section about its axis $O$ : the inner electrode of radius $R_1$ , the layer $\varepsilon_1$ out to $R_2$ , the layer $\varepsilon_2$ out to the outer electrode at $R_3$ .

Given

radii $R_1 < R_2 < R_3$
layers $\varepsilon_1$ (inner), $\varepsilon_2$ (outer)
breakdown fields $E_1$ , $E_2$
$\varepsilon_1 R_1 E_1 < \varepsilon_2 R_2 E_2$

Find

the breakdown voltage

V

Solution

Step 1The field in the two layers

Let the inner electrode carry the charge $\lambda$ per unit length. Far from the ends, by symmetry $\vect D$ is radial and the same all round the axis, and Gauss’s theorem on a coaxial cylinder of radius $r$ and length $\ell$ gives

D\cdot 2\pi r\ell = \lambda\ell

Dividing by $\varepsilon_0\varepsilon$ of each layer:

\begin{gathered} E(r) = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_1 r}\quad (R_1 < r < R_2) \\ E(r) = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_2 r}\quad (R_2 < r < R_3) \end{gathered} \tag{1}

The field with the inner electrode positive, drawn for ε_1 = 2, ε_2 = 3: the outer layer carries fewer lines, and the rest end on the negative bound charge at the boundary R_2. — The field with the inner electrode positive, drawn for $\varepsilon_1 = 2$ , $\varepsilon_2 = 3$ : the outer layer carries fewer lines, and the rest end on the negative bound charge at the boundary $R_2$ .

Step 2Which layer breaks down first

In each layer the field falls as $1/r$ , so its largest value is at the layer’s inner surface. By (1) the two layers reach breakdown at the charges

\begin{gathered} \lambda_1 = 2\pi\varepsilon_0\,\varepsilon_1 R_1 E_1 \quad (\text{at } r = R_1) \\ \lambda_2 = 2\pi\varepsilon_0\,\varepsilon_2 R_2 E_2 \quad (\text{at } r = R_2) \end{gathered}

The given inequality $\varepsilon_1 R_1 E_1 < \varepsilon_2 R_2 E_2$ says that $\lambda_1 < \lambda_2$ . As the voltage rises, $\lambda$ reaches $\lambda_1$ first, so the inner layer breaks down first, at its surface next to the inner electrode, and the capacitor fails there:

\lambda_{\mathrm{b}} = 2\pi\varepsilon_0\,\varepsilon_1 R_1 E_1 \tag{2}

Step 3The voltage at that charge

The voltage is the line integral of (1) across both layers:

\begin{aligned} V &= \int_{R_1}^{R_2}\frac{\lambda\,dr}{2\pi\varepsilon_0\varepsilon_1 r} + \int_{R_2}^{R_3}\frac{\lambda\,dr}{2\pi\varepsilon_0\varepsilon_2 r}\\ &= \frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{\varepsilon_1}\ln\frac{R_2}{R_1} + \frac{1}{\varepsilon_2}\ln\frac{R_3}{R_2}\right) \end{aligned}

Setting $\lambda$ to (2), the factor $2\pi\varepsilon_0$ cancels and $\varepsilon_1$ multiplies through the bracket:

V = R_1 E_1\left[\ln\dfrac{R_2}{R_1} + \dfrac{\varepsilon_1}{\varepsilon_2}\ln\dfrac{R_3}{R_2}\right]

(3)

Check

Dimensions: $R_1 E_1$ is a length times a field, a voltage, and the bracket is a pure number. Single layer: with $\varepsilon_2 = \varepsilon_1$ the bracket becomes $\ln(R_3/R_1)$ and $V = R_1 E_1\ln(R_3/R_1)$ , the breakdown voltage of a uniformly filled cylindrical capacitor whose peak field $V/(R_1\ln(R_3/R_1))$ sits at the inner electrode. And with $R_3 = R_2$ (no outer layer) the result reduces to $R_1E_1\ln(R_2/R_1)$ in the same way.

Remark

Only the inner layer’s strength $E_1$ enters the answer; $E_2$ matters only through the inequality that picks which layer fails. A higher permittivity $\varepsilon_2$ outside lowers $V$ , since the outer layer then takes a smaller share of the voltage and more of it falls on the inner layer, which is already the weak one. When the inequality becomes an equality, both layers fail at the same voltage and neither is stronger than it needs to be.