Dielectrics and Gauss' Law — Question 2

Question 2

An oblique dielectric interface preserves tangential field and normal displacement

Two homogeneous linear dielectrics meet at a flat interface with no free surface charge. Let n point from medium 1 to 2 and t be tangent. Relative permittivities are 2 and 5; $\mathbf E_1=(3\hat{\mathbf n}+4\hat{\mathbf t})$ V/m.

1. Find E2 using $D_{2n}=D_{1n}$ and $E_{2t}=E_{1t}$.

2. Find the net bound interface charge density using $\sigma_b=(\mathbf P_1-\mathbf P_2)\cdot\hat{\mathbf n}$.

Question 2 - Solution

An oblique dielectric interface preserves tangential field and normal displacement

(a) Method and calculation

$$2\epsilon_0(3)=5\epsilon_0E_{2n}\Rightarrow E_{2n}=1.20\,\mathrm{V/m}.$$$$\mathbf E_2=(1.20\hat{\mathbf n}+4.00\hat{\mathbf t})\,\mathrm{V/m}.$$

(b) Interpretation and further calculation

$$P_{1n}=\epsilon_0(2-1)(3)=3\epsilon_0\,\mathrm{V/m},\quad P_{2n}=\epsilon_0(5-1)(1.20)=4.80\epsilon_0\,\mathrm{V/m}.$$$$\sigma_b=-1.80\epsilon_0\,\mathrm{V/m},\quad E_{2n}-E_{1n}=\sigma_b/\epsilon_0.$$

Physical check

A jump in normal E is allowed even without free interface charge because bound interface charge is generally present.