Question

a set of charged plates have an area of 2.22*10^-m^2. When 5.24*10^-9 C of charge in placed on the plates, it creates a potential difference of 240 V. What is the separation between the plates?

Answer 1

$$q=CV\\q=\frac{\epsilon_0 A}{d}V\\(5.24\cdot10^{-9})=\frac{(2.22\cdot10^{-4})(8.85\cdot10^{-12})}{d}240\\d=9\cdot10^{-5}\ m$$