Dielectrics and Gauss' Law — Question 7

Question 7

A prescribed polarization profile changes voltage through its spatial average

Use $\mathbf D=\epsilon_0\mathbf E+\mathbf P$, $\nabla\cdot\mathbf D=\rho_f$, $\rho_b=-\nabla\cdot\mathbf P$ and $\sigma_b=\mathbf P\cdot\hat{\mathbf n}$, where n points out of the dielectric. In a plate gap $0<z<d$, the prescribed polarization is $\mathbf P=P_0(z/d)^2\hat{\mathbf z}$. The electrodes carry free densities $+\sigma_f$ at zero and $-\sigma_f$ at d. There is no free bulk charge.

1. Find the bound volume and surface charge densities and verify their net sum is zero.

2. Find E(z) and the voltage drop from the lower electrode to the upper.

Question 7 - Solution

A prescribed polarization profile changes voltage through its spatial average

(a) Method and calculation

$$\rho_b=-2P_0z/d^2,\quad\sigma_b(0)=0,\quad\sigma_b(d)=P_0.$$$$\int_0^d\rho_b\,dz+\sigma_b(0)+\sigma_b(d)=-P_0+P_0=0.$$

(b) Interpretation and further calculation

$$D_z=\sigma_f\Rightarrow E_z=\frac{\sigma_f-P_0(z/d)^2}{\epsilon_0}.$$$$V_{\mathrm{lower}}-V_{\mathrm{upper}}=\int_0^dE_zdz=\frac d{\epsilon_0}(\sigma_f-P_0/3).$$

Physical check

This is a prescribed polarization model; it need not obey a constant-permittivity constitutive relation.