Insulators — Question 7

Question 7

Series dielectric layers concentrate the electric field in the lower permittivity

Two planar insulating layers in series have thicknesses $d_1=d_2=d$ and permittivities $\varepsilon_1=2\varepsilon_0$, $\varepsilon_2=6\varepsilon_0$. There is no free interface charge. Apply total voltage V normal to the layers.

1. Find both fields in terms of V and d.

2. If both materials have breakdown field $E_b$, find the total voltage at first breakdown and both voltage fractions.

Question 7 - Solution

Series dielectric layers concentrate the electric field in the lower permittivity

(a) Method and calculation

$$D=\varepsilon_1E_1=\varepsilon_2E_2\Rightarrow E_1=3E_2.$$$$V=d(E_1+E_2)\Rightarrow\boxed{E_1=3V/(4d),\quad E_2=V/(4d)}.$$

(b) Interpretation and further calculation

$$E_1=E_b\Rightarrow\boxed{V_{\max}=4dE_b/3}.$$$$\boxed{V_1/V=3/4,\quad V_2/V=1/4,\quad E_2=E_b/3\text{ at first breakdown}}.$$

Physical check

Continuity applies to normal displacement D in the absence of free interface charge, not to E. The lower-permittivity layer reaches the common breakdown field first.