Dielectrics: An Atomic View — Question 10

Question 10

Saturation separates large-signal permittivity from incremental capacitance

A dielectric has $P(E)=P_s\tanh(E/E_s)$ with Ps,Es positive. It fills a plate capacitor of area A and gap d. Use $Q=A[\epsilon_0E+P(E)]$ and V=Ed.

1. Find the secant capacitance Q/V for nonzero E and the differential capacitance dQ/dV.

2. Find their zero-field limits and their limits as positive E becomes very large.

Question 10 - Solution

Saturation separates large-signal permittivity from incremental capacitance

(a) Method and calculation

$$C_s(E)=\frac Ad\left[\epsilon_0+\frac{P_s}{E}\tanh(E/E_s)\right].$$$$C_d(E)=\frac Ad\left[\epsilon_0+\frac{P_s}{E_s}\operatorname{sech}^2(E/E_s)\right].$$

(b) Interpretation and further calculation

$$E\to0:\quad C_s,C_d\to\frac Ad(\epsilon_0+P_s/E_s).$$$$E\to+\infty:\quad C_s,C_d\to\epsilon_0A/d,\quad P\to P_s.$$

Physical check

Polarization saturates while the vacuum contribution to electrode charge keeps increasing; the incremental material response tends to zero.