Source: High school physics (Chinese)
Problem
An air-filled parallel-plate capacitor has plate separation $d = 1.0$ mm. Both plates are squares of equal area.
- What side length must each square have for a capacitance of $100$ pF?
- What side length must each square have for a capacitance of $1.0\ \mu\mathrm{F}$?
- What side length must each square have for a capacitance of $1.0$ F?
(1) $L \approx 11$ cm. (2) $L \approx 10.6$ m. (3) $L \approx 10.6$ km.
For an air parallel-plate capacitor: $C = \dfrac{\varepsilon_0 A}{d}$, where $A = L^2$ for square plates of side $L$. Solving for $L$:
$$L = \sqrt{\dfrac{Cd}{\varepsilon_0}}.$$With $d = 1.0 \times 10^{-3}$ m and $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m:
(1) $C = 100$ pF $= 1.0 \times 10^{-10}$ F: $L = \sqrt{\dfrac{1.0 \times 10^{-10} \times 1.0 \times 10^{-3}}{8.85 \times 10^{-12}}} \approx 0.106$ m $\approx 11$ cm.
(2) $C = 1.0\ \mu\mathrm{F} = 1.0 \times 10^{-6}$ F: $L \approx 10.6$ m.
(3) $C = 1.0$ F: $L \approx 1.06 \times 10^{4}$ m $\approx 10.6$ km.
The dramatic scaling shows why parallel plates with air dielectric cannot practically deliver microfarad-or-larger capacitances --- real high-$C$ capacitors use thin dielectrics with high $\varepsilon_r$ in rolled or stacked geometry.