Consider the following parallel-plate capacitor, with a potential difference of $V$ across it's plates:

enter image description here

I've seen some problems that assume a dielectric with a variable electric permittivity of $$\epsilon=\epsilon_0\left(1+{z\over a} \right)$$ or something similar (but a function of z) between the plates.

With these variable permittivities, how can we, as usual, apply boundary conditions and also use the potential formula, that we use to solve dielectric problems?:

$$V=\int_{z=0}^{z=a} \vec{E}.d \vec l$$ $$\epsilon_{above}\vec{E}_{above}.\hat n-\epsilon_{below}\vec{E}_{below}.\hat n=\sigma_{free}$$ or equivalently, how can we find the electric field between the plates?

  • $\begingroup$ Try representing your dielectric as consisting of $n$ layers of different (but constant inside layer) permittivity. Then use usual boundary conditions between these layers, and take the limit $n\to\infty$ to get final equation. $\endgroup$ – Ruslan Aug 4 '13 at 18:43

Everywhere inside of the dielectric, the following (Gauss's Law inside of meadia) equation holds $$ \nabla\cdot\mathbf D = \rho_\mathrm{free}, \qquad \mathbf D = \epsilon\mathbf E $$ Inside of the dielectric, there is no free charge, so we have the equation $$ \nabla\cdot(\epsilon\mathbf E) = 0, \qquad 0<z<a $$ Now, we recall the definition of the electric potential; $$ \mathbf E = -\nabla V $$ which therefore gives $$ \nabla\cdot(\epsilon\nabla V) = 0, \qquad 0<z<a $$ the problem is symmetric with respect to rotations around $z$, so we take an ansatz $$ V(x,y,z) = v(z) $$ then the above equation gives $$ \frac{d}{dz}\left(\epsilon(z) \frac{d}{dz}v(z)\right) = 0, \qquad 0<z<a $$ Now, you just need to solve this equation subject to the appropriate boundary conditions $$ v(0) = 0, \qquad v(a) = V_0 $$ Once you have the potential, you can get the electric field by taking the gradient.

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    $\begingroup$ Thanks a lot! One point: It seems the final DE should be ${d\over dz}(\epsilon (z) {dv(z)\over dz})=0$. Is that right ? $\endgroup$ – Zorich Aug 4 '13 at 19:46
  • $\begingroup$ And if my equation in the comment be the correct equation, then I can not find the field, because I don't have the needed boundary conditions for ${dv\over dz}$. $\endgroup$ – Zorich Aug 4 '13 at 20:04
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    $\begingroup$ Yes that was a typo thanks. The resulting second order ordinary differential equation can be solved using your two boundary data. You do not need the normal derivative of the potential at the boundary. Integrate the equation once, then use separation of variables and then integrate again. You'll get two undetermined constants in the resulting expression for $v(z)$ that can be solved for using the boundary data. $\endgroup$ – joshphysics Aug 4 '13 at 20:08

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