Capacitor with a Dielectric — Question 8

Question 8

A graded dielectric requires a harmonic spatial average of permittivity

A plate gap $0\le z\le d$ is filled with $\kappa(z)=1+\alpha z/d$, where alpha>0. No free charge lies within the dielectric, so D is constant and $E(z)=D/[\epsilon_0\kappa(z)]$.

1. Find the capacitance from the voltage integral.

2. Find the field ratio at the two faces and the limit as alpha tends to zero.

Question 8 - Solution

A graded dielectric requires a harmonic spatial average of permittivity

(a) Method and calculation

$$V=\frac D{\epsilon_0}\int_0^d\frac{dz}{1+\alpha z/d}=\frac{Dd}{\epsilon_0\alpha}\ln(1+\alpha).$$$$C=\frac{DA}{V}=\frac{\epsilon_0A}{d}\frac{\alpha}{\ln(1+\alpha)}.$$

(b) Interpretation and further calculation

$$E(0)/E(d)=1+\alpha.$$$$\lim_{\alpha\to0}\frac{\alpha}{\ln(1+\alpha)}=1\Rightarrow C\to\epsilon_0A/d.$$

Physical check

The lowest-permittivity part has the largest field and contributes most voltage per unit thickness.