Irodov 3.106: Two Weak Spots Failing Together – Amit & Deepali Sinha Foundation

Problem Irodov 3.106

The gap of a cylindrical capacitor is filled with two coaxial cylindrical layers of dielectric, of permittivities $\varepsilon_1$ and $\varepsilon_2$ , whose inner radii are $R_1$ and $R_2$ respectively, with $R_2 > R_1$ . The largest field each dielectric can withstand is $E_{1m}$ for the first and $E_{2m}$ for the second. As the voltage across the capacitor is raised, what relation between the permittivities, the radii and the breakdown fields makes the field reach the breakdown value in both dielectrics at the same moment?

The capacitor end on: the inner electrode of radius R_1 about the axis O, the layer ε_1 from R_1 to R_2, and the layer ε_2 from R_2 out to the outer electrode. — The capacitor end on: the inner electrode of radius $R_1$ about the axis $O$ , the layer $\varepsilon_1$ from $R_1$ to $R_2$ , and the layer $\varepsilon_2$ from $R_2$ out to the outer electrode.

Given

layers $\varepsilon_1$ (inner radius $R_1$ ) and $\varepsilon_2$ (inner radius $R_2 > R_1$ )
breakdown fields $E_{1m}$ , $E_{2m}$

Find

the relation for simultaneous breakdown

Solution

Step 1The field in each layer

Let the inner electrode carry the charge $\lambda$ per unit length. Far from the ends, the field is radial and the same all round the axis, so Gauss’s theorem for $\vect D$ on a coaxial cylinder of radius $r$ and length $\ell$ sees only the free charge:

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

Dividing $D$ by $\varepsilon_0\varepsilon$ of the layer the radius lies in gives the field in each layer:

\begin{gathered} E_1(r) = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_1 r}\quad (R_1 < r < R_2) \\ E_2(r) = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_2 r}\quad (r > R_2) \end{gathered} \tag{1}

Step 2Where each layer is most stressed

The field against the distance from the axis, drawn for ε_1 = 2, ε_2 = 1.25. It falls as 1/r in each layer and jumps at the boundary R_2; each layer's largest field sits at its inner surface (R_3 is the outer electrode). — The field against the distance from the axis, drawn for $\varepsilon_1 = 2$ , $\varepsilon_2 = 1.25$ . It falls as $1/r$ in each layer and jumps at the boundary $R_2$ ; each layer’s largest field sits at its inner surface ( $R_3$ is the outer electrode).

In each layer the field (1) falls as $1/r$ , so it is largest at the layer’s inner surface, $r = R_1$ for the first layer and $r = R_2$ for the second:

\begin{gathered} E_{1,\mathrm{max}} = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_1 R_1} \\ E_{2,\mathrm{max}} = \frac{\lambda}{2\pi\varepsilon_0\varepsilon_2 R_2} \end{gathered} \tag{2}

Breakdown in a layer starts where its field first reaches the permissible value, so these two maxima are the ones to watch.

Step 3Both maxima reach breakdown together

Raising the voltage raises $\lambda$ and both maxima in (2) in the same proportion. Both dielectrics fail at the same voltage exactly when the same $\lambda$ brings each maximum to its limit:

\begin{gathered} \frac{\lambda}{2\pi\varepsilon_0\varepsilon_1 R_1} = E_{1m} \\ \frac{\lambda}{2\pi\varepsilon_0\varepsilon_2 R_2} = E_{2m} \end{gathered}

Solving each for $\lambda/(2\pi\varepsilon_0)$ and equating the two values:

\varepsilon_1 R_1 E_{1m} = \varepsilon_2 R_2 E_{2m}

(3)

If instead $\varepsilon_1 R_1 E_{1m} < \varepsilon_2 R_2 E_{2m}$ , the inner layer breaks down first, and in the opposite case the outer one does.

Check

Each side of (3) is a permittivity (a pure number) times a length times a field, so both have the same unit, volts. Special case: one material throughout, $\varepsilon_1 = \varepsilon_2$ and $E_{1m} = E_{2m}$ ; then the condition demands $R_1 = R_2$ , so the two “layers” share the same inner surface, the only place a single material is stressed to its maximum. With $R_2 > R_1$ the outer part of a uniform filling never reaches breakdown before the inner part, as (1) says.

Remark

By (2), $2\pi\varepsilon_0\varepsilon R E_m$ is the largest charge per unit length the inner electrode can carry before that layer breaks down, so the condition says both layers hold out to the same charge. This is the principle of graded insulation in high-voltage cables: putting the material of higher permittivity next to the core lowers the field there and evens out the stress, so that no part of the insulation is overdesigned.