Dielectrics and Gauss' Law — Question 1

Question 1

Gauss law separates electrode charge from the bound charge that screens it

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. A linear dielectric of relative permittivity kappa fills a plate gap from z=0 to d. The electrode at zero has free charge density $+\sigma_f$ and the one at d has $-\sigma_f$, with $\sigma_f>0$. Neglect fringing.

1. Find D, E and P in the gap.

2. Find the bound charge on each dielectric face and verify the field using total charge.

Question 1 - Solution

Gauss law separates electrode charge from the bound charge that screens it

(a) Method and calculation

$$\mathbf D=\sigma_f\hat{\mathbf z},\quad\mathbf E=\frac{\sigma_f}{\kappa\epsilon_0}\hat{\mathbf z}.$$$$\mathbf P=\epsilon_0(\kappa-1)\mathbf E=\sigma_f(1-1/\kappa)\hat{\mathbf z}.$$

(b) Interpretation and further calculation

$$\sigma_b(0)=-\sigma_f(1-1/\kappa),\quad\sigma_b(d)=+\sigma_f(1-1/\kappa),\quad\rho_b=0.$$$$E=\frac{\sigma_f+\sigma_b(0)}{\epsilon_0}=\frac{\sigma_f}{\kappa\epsilon_0}.$$

Physical check

Gauss law for D counts free charge; Gauss law for E counts free plus bound charge.