Capacitor with a Dielectric — Question 6

Question 6

The first dielectric layer to reach its field ceiling sets the total voltage limit

Use $\epsilon_0=8.854\times10^{-12}$ F/m. Two equal-thickness layers fill a total gap d=1.00 mm. Their relative permittivities are 2 and 5, and their supplied field ceilings are 3.00 and 1.00 MV/m respectively. No free interface charge is present.

1. Find the greatest allowed common displacement magnitude $D=\epsilon_iE_i$.

2. Find each layer field and the maximum applied voltage.

Question 6 - Solution

The first dielectric layer to reach its field ceiling sets the total voltage limit

(a) Method and calculation

$$D\le\epsilon_0\min(2\times3.00,5\times1.00)\times10^6.$$$$D_{\max}=5.00\times10^6\epsilon_0=44.2700\,\mu\mathrm{C/m^2}.$$

(b) Interpretation and further calculation

$$E_1=D_{\max}/(2\epsilon_0)=2.50\,\mathrm{MV/m},\quad E_2=1.00\,\mathrm{MV/m}.$$$$V_{\max}=(d/2)(E_1+E_2)=1750\,\mathrm V.$$

Physical check

The second material limits the device even though its permittivity is larger; multiplying total thickness by just one field ceiling is incorrect.