0
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ What is D supposed to be? If there is surface charge then D is discontinuous. $\endgroup$
    – ProfRob
    Commented May 6, 2020 at 22:37
  • $\begingroup$ If your surface is $z=-1$, whose normal is $\hat{z}$, then how is $\vec{E}$ along $\hat{x}$? $\endgroup$
    – Vivek
    Commented May 6, 2020 at 22:41
  • $\begingroup$ I'm talking about discontinuity of electric field across the boundry of dielectric slab and free space. Edited question a bit to add versors. $\endgroup$
    – zakrent
    Commented May 6, 2020 at 22:47
  • $\begingroup$ Isn't this the typical example of showing the dangers of applying Gauss's law for the electric displacement across a discontinuous boundary? $\endgroup$ Commented May 6, 2020 at 22:50
  • 1
    $\begingroup$ 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. $\endgroup$ Commented May 6, 2020 at 23:10

3 Answers 3

0
$\begingroup$

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 $\vec{E}.$

$\endgroup$
4
  • $\begingroup$ I'm talking about discontinuity of electric field across the boundry of dielectric slab ($x < a, z > -1$) and free space. $\endgroup$
    – zakrent
    Commented May 6, 2020 at 22:49
  • $\begingroup$ 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. $\endgroup$
    – sslucifer
    Commented May 6, 2020 at 22:57
  • $\begingroup$ Dielectric doesn't have any free charges so it shouldn't affect the D field, is the equation for E field incorrect? $\endgroup$
    – zakrent
    Commented May 6, 2020 at 23:12
  • $\begingroup$ 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. $\endgroup$
    – sslucifer
    Commented May 6, 2020 at 23:17
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

The tangential component of the electric field across the interface at $x=a$ is continuous, so your expressions for the E-field are incorrect.

If there is an intersection of a charged plane with a dielectric, then you cannot assume there is no field tangential to the charged plane as you would for a plane in a vacuum. There will be a tangential component to the conducting plane (i.e. a normal component to the dielectric plane) because the symmetry of the usual infinite charged plane problem is broken.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.