# Gauss Law - tangential component discontinuity at dielectric boundary

Let's say we have a dielectric slab $$x < a, z > -1$$ and a plane $$z=-1$$ with a surface charge of $$\sigma$$. Gauss law states that: $$\oint_S D dA = Q_{free}$$ Dielectric has no free charge so electric displacement field can be expressed as electric displacement field of a plane: $$D = \frac{\sigma}{2}$$ Where $$\vec{D}$$ points in positive $$z$$ direction for $$z > -1$$. Then electric field along the x axis can be expressed as: $$\vec{E(x)} = \left\{\begin{array}{ll} \frac{\sigma}{2\epsilon_0}\hat{z} & \textrm{for x > a }\\ \frac{\sigma}{2\epsilon_0\epsilon_r}\hat{z} & \textrm{for x < a }\\ \end{array}\right.$$ Which causes a discontinuity at $$x = a$$ at the surface of a slab $$x < a, z > -1$$ but tangental component of $$E$$ field should be continious across the interface. What part of electrostatic theory am I missing?

• What is D supposed to be? If there is surface charge then D is discontinuous. Commented May 6, 2020 at 22:37
• If your surface is $z=-1$, whose normal is $\hat{z}$, then how is $\vec{E}$ along $\hat{x}$? Commented May 6, 2020 at 22:41
• I'm talking about discontinuity of electric field across the boundry of dielectric slab and free space. Edited question a bit to add versors. Commented May 6, 2020 at 22:47
• Isn't this the typical example of showing the dangers of applying Gauss's law for the electric displacement across a discontinuous boundary? Commented May 6, 2020 at 22:50
• I don't have time to type out an answer. But if you have access to Griffith's Introduction to Electrodynamics, check out chapter 4. Commented May 6, 2020 at 23:10

So there is two surface $$x=a$$ and $$z=-1$$, so the surface integral can't simply be ordinary integral and in that way it is not that straight forward expression of $$\vec{E}.$$
• I'm talking about discontinuity of electric field across the boundry of dielectric slab ($x < a, z > -1$) and free space. Commented May 6, 2020 at 22:49
• Oh yes I see, then you can't simply evaluate the Gauss integral, as there is two surfaces. One $x=a$ and other $z=-1$. So the simplification of surface integral into proper integral will not be valid. Commented May 6, 2020 at 22:57
• D is also a vector quantity, so you should figure out what is the direction of that vector. I think you can't or it is difficult because of the surface integral. The expression of $\vec{E}$ is correct, if $\vec{D}=\sigma/2\hat{z}$, but it think its not. $\sigma/2\hat{z}$ is when you have an infinite plane $x=a$, but its not. Commented May 6, 2020 at 23:17
I think your equations for $$\bf E$$ do not apply at the edge the dielectric in this situation, owing to the polarization being in a direction different to $$\bf D$$ near the edge. However I have not worked it out fully so am not quite sure.
The tangential component of the electric field across the interface at $$x=a$$ is continuous, so your expressions for the E-field are incorrect.