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