Electrostatics
Intermediate
Capacitor
Source: High school physics (Chinese)
Problem
Design a capacitor of capacitance $0.1\ \mu\mathrm{F}$ that operates safely at $1000$ V. The dielectric has relative permittivity $\varepsilon_r = 3$ and breakdown field strength $E_{\text{breakdown}} = 10$ kV/mm. Use a safety factor of $4$ on the breakdown field.
Find the required plate area.
$d \approx 0.4$ mm, $A \approx 1.5\ \mathrm{m}^2$.
With safety factor $4$, the maximum operating field is $E_{\text{op}} = E_{\text{breakdown}} / 4 = 2.5$ kV/mm $= 2.5 \times 10^{6}$ V/m.
The required dielectric thickness is set by the operating voltage:
$$d = \dfrac{V}{E_{\text{op}}} = \dfrac{1000}{2.5 \times 10^{6}} = 4.0 \times 10^{-4}\ \mathrm{m} = 0.4\ \mathrm{mm}.$$The plate area follows from $C = \varepsilon_r \varepsilon_0 A / d$:
$$A = \dfrac{C d}{\varepsilon_r \varepsilon_0} = \dfrac{(0.1 \times 10^{-6})(4.0 \times 10^{-4})}{3 \times 8.85 \times 10^{-12}} \approx 1.5\ \mathrm{m}^2.$$