Capacitance of a Parallel-Plate Capacitor with Graded Permittivity: 04-BS-9, December 2018, Q1

Question 1: Capacitance of a Parallel-Plate Capacitor with Graded Permittivity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Plate separation $d$	$1\times10^{-3}$ m
Plate area $A$ (circular)	$5\ \text{cm}^2=5\times10^{-4}\ \text{m}^2$
Permittivity profile	$\varepsilon(z)=\varepsilon_1/(1+\alpha z/2)$
$\varepsilon_1$, $\alpha$	$3$, $0.2/d$

Find. The capacitance $C$ of the graded-dielectric capacitor.

Graded-dielectric capacitor: permittivity is highest (darker shading) at the z=0 plate and decreases toward the z=d plate. D is uniform and normal to the plates throughout, so the gap behaves as a stack of series capacitors.

Approach. With no free charge inside the dielectric, the displacement field $D$ is uniform and normal to the plates, so the gap is equivalent to a continuum of infinitesimally thin capacitors in series — integrate the reciprocal permittivity through the gap.

Set up the series-capacitor integral. A thin slab of thickness $dz$ at position $z$ behaves as a capacitor $dC=\varepsilon_0\varepsilon(z)A/dz$; stacking these in series along $z$ gives $$\frac{1}{C}=\int_0^d\frac{dz}{\varepsilon_0\varepsilon(z)A}=\frac{1}{\varepsilon_0A\varepsilon_1}\int_0^d\left(1+\frac{\alpha z}{2}\right)dz$$ using $1/\varepsilon(z)=(1+\alpha z/2)/\varepsilon_1$.
Evaluate the integral. $$\int_0^d\left(1+\frac{\alpha z}{2}\right)dz=d+\frac{\alpha d^2}{4}$$ Substituting $\alpha=0.2/d$ gives $\alpha d^2/4=0.05d$, so the integral equals $1.05d$: $$\frac{1}{C}=\frac{1.05\,d}{\varepsilon_0A\varepsilon_1}\quad\Rightarrow\quad C=\frac{\varepsilon_0A\varepsilon_1}{1.05\,d}$$
Substitute the numbers. $\varepsilon_0=8.85\times10^{-12}$ F/m, $A=5\times10^{-4}$ m$^2$, $\varepsilon_1=3$, $d=1\times10^{-3}$ m: $$C=\frac{(8.85\times10^{-12})(5\times10^{-4})(3)}{1.05\times10^{-3}}$$ $$\boxed{C=1.264\times10^{-11}\ \text{F}=12.64\ \text{pF}}$$

Quantity	Result
Series-integral factor (vs. plain $\varepsilon_1$ formula)	$1.05$
Capacitance $C$	$1.264\times10^{-11}$ F $=12.64$ pF

Concept & Theory

Topic. Series-capacitor model for a graded (spatially-varying) dielectric.

Key relations. No free charge inside the dielectric $\Rightarrow D$ continuous and uniform through the gap (Gauss's law); $C_{\text{series}}=\left[\displaystyle\int dz/(\varepsilon_0\varepsilon(z)A)\right]^{-1}$.

Why this works. All field lines run straight from plate to plate (normal incidence, no free charge inside), so slicing the gap into infinitesimal parallel slabs and combining them as series capacitors is exact — it is the direct continuum generalisation of the uniform parallel-plate formula.

Common pitfall. Integrating $\varepsilon(z)$ itself (as if computing an average permittivity) instead of integrating $1/\varepsilon(z)$ — capacitors in series combine as reciprocals, exactly like resistors in series combine directly while conductances combine as reciprocals.

Source. Sadiku, Elements of Electromagnetics, Ch. 5–6 (dielectrics, boundary conditions, capacitance).

Question 1 of 8: Capacitance of a Parallel-Plate Capacitor with Graded Permittivity

Question 1: Capacitance of a Parallel-Plate Capacitor with Graded Permittivity (20 marks)