Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Exercises - Page 593: 37

$0.418\,mm$

Given/known: $C=9.00\,nF=9.00\times10^{-9}\,F,$ $A=0.425\,m^{2}$ and $\varepsilon_{0}=8.854\times10^{-12}\,F/m$. Recall: Capacitance C(in air)=$\frac{\varepsilon_{0}A}{d}$ or $d=\frac{\varepsilon_{0}A}{C}$ Result: $d=\frac{8.854\times10^{-12}\,F/m\times0.425\,m^{2}}{9.00\times10^{-9}\,F}=0.418\times10^{-3}\,m$ $=0.418\,mm$