= Solution
Write the plate area as $S$ and separation as $d$, so the original <capacitance> is $C=\varepsilon_0S/d$. Neglect edge fields. The greatest <capacitance> occurs when each material occupies half the plate area and spans the full separation. These two regions are <capacitors> in parallel, sharing the same <electric potential difference>:
$$
\boxed{C_{\max}=\frac C2+\frac{\epsilon C}{2}
=\frac{1+\epsilon}{2}C.}
$$
The least <capacitance> occurs when a dielectric layer of thickness $d/2$ covers the whole area, leaving a vacuum layer of the same thickness. The two layers are <capacitors> in series: the normal <electric displacement field> is continuous and the voltage drops add. Their individual capacitances are $2\epsilon C$ and $2C$, giving
$$
\boxed{C_{\min}=\left(\frac1{2C}+\frac1{2\epsilon C}\right)^{-1}
=\frac{2\epsilon}{1+\epsilon}C.}
$$
The <arithmetic-harmonic bounds for dielectric capacitance> show that these are genuine extrema over other arrangements of the same two volumes, within the ideal parallel-plate model. A uniform trial <electric field> in the fixed-voltage energy principle bounds $C$ above by the arithmetic mean of the two permittivities; a uniform trial <electric displacement field> in the fixed-charge principle bounds it below by their harmonic mean. The two layer arrangements attain those bounds. Both reduce to $C$ when $\epsilon=1$.
Back to article page
