Electrostatics
Intermediate
Capacitor
Source: High school physics (Chinese)
Problem
The breakdown field strength of dry air is $3 \times 10^{3}$ V/m. A parallel-plate capacitor has plates of area $40 \times 50\ \mathrm{cm}^2$.
What is the maximum charge this capacitor can store?
$Q_{\max} \approx 5.3 \times 10^{-9}$ C $\approx 5.3$ nC.
At the breakdown field, the surface charge density is $\sigma_{\max} = \varepsilon_0 E_{\text{breakdown}}$, so the maximum stored charge is:
$$Q_{\max} = \sigma_{\max} A = \varepsilon_0 \, A \, E_{\text{breakdown}}.$$(Note that $d$ cancels: $Q = CV = (\varepsilon_0 A / d)(E d) = \varepsilon_0 A E$.)
With $A = 40 \times 50\ \mathrm{cm}^2 = 0.20\ \mathrm{m}^2$, $E_{\text{breakdown}} = 3 \times 10^{3}$ V/m, and $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m:
$$Q_{\max} = 8.85 \times 10^{-12} \times 0.20 \times 3 \times 10^{3} \approx 5.3 \times 10^{-9}\ \mathrm{C}.$$