Variable permittivity capacitor (Homework 3-310)

The capacitor is made of the two parallel plates of sizes $a$ and $b$, along the $x$ and $y$ axes, respectively, with the distance $d$ between the plates.  The empty space between the plates is filled by a non-uniform dielectric. Determine the capacitance of this capacitor, neglecting end effects, if the dielectric function of the dielectric is given by

a) $epsilon(x)=epsilon_0(2-cos(4pi x/a))$

b) $epsilon(x)=epsilon_0(1+x/a)$

c) $epsilon(x)=epsilon_0(1+x^2/a^2)$

Homework Capacitor

Between the plates of a parallel plate capacitor the electric field is given by

$E= (1/epsilon)*(Q/S)=(1/epsilon)*(Q/ab)$

Since $epsilon=epsilon(x)$ it means $E=E(x)$  varies along the length x of the capacitor (still constant along z. For the infinitesimal surface in the figure the field between plates is

$E(x)=frac{Q}{d x*b*epsilon(x)}$

The difference of potential between the plates varies also along the length x of the capacitor

$V(z)=-int_{z_1}^{z_2}E(x)dz=frac{Q d}{b*d x*epsilon(x)}$

Therefore for the infinitesimal stripe of length $d x$ the capacitance is

$d C(x)=frac{Q}{V(x)}=frac{b*d x*epsilon(x)}{d}$

All the stripes in the figures are connected in parallel so that their capacitance add all together:

$C_{tot}=int d C(x)=frac{b}{d}int_{0}^{a}epsilon(x)d x$

In case a) one has

$C_{tot}=frac{b}{d}int_{0}^{a}left[2-cos(frac{4pi x}{a}) right]d x=frac{b}{d}left[2x-(a/4pi)sin(frac{4pi x}{a}) right]|_0^a=epsilon_0*frac{b}{d}*2a$

In case b) one has

$C_{tot}=epsilon_0*frac{b}{d}int_{0}^{a}(1+frac{x}{a})d x=epsilon_0*frac{b}{d}*(x+frac{x^2}{2a})|_0^2a=epsilon_0*frac{b}{d}*frac{3a}{2}$

In case c) one has

$C_{tot}=epsilon_0*frac{b}{d}int_{0}^{a}(1+frac{x^2}{a^2})d x=epsilon_0*frac{b}{d}*(x+frac{x^3}{3a^2})|_0^a=epsilon_0*frac{b}{d}*frac{4a}{3}$