Dielectrics and Gauss' Law — Question 3

Question 3

A free charge sheet makes displacement different on the two sides of a layered gap

Two dielectric layers of thicknesses $t_1,t_2$ and permittivities $\epsilon_1,\epsilon_2$ fill a plate gap. A fixed free interface sheet has density $\sigma_s$. Let z run from the lower electrode to the upper, and define V as lower minus upper potential. Fringing is negligible.

1. Find D1 and D2 using the voltage and the interface jump.

2. Find the two electrode free charge densities and verify charge neutrality including the interface sheet.

Question 3 - Solution

A free charge sheet makes displacement different on the two sides of a layered gap

(a) Method and calculation

$$D_2-D_1=\sigma_s,\quad V=D_1t_1/\epsilon_1+D_2t_2/\epsilon_2.$$$$D_1=\frac{V-\sigma_st_2/\epsilon_2}{t_1/\epsilon_1+t_2/\epsilon_2},\quad D_2=D_1+\sigma_s.$$

(b) Interpretation and further calculation

$$\sigma_{f,\mathrm{lower}}=D_1,\quad\sigma_{f,\mathrm{upper}}=-D_2.$$$$\sigma_{f,\mathrm{lower}}+\sigma_s+\sigma_{f,\mathrm{upper}}=D_1+\sigma_s-D_2=0.$$

Physical check

Opposite electrode charges need not have equal magnitudes when free charge is trapped inside the gap.