Question

Calculate the potential difference of a battery connected to a capacitor consisting of two parallel plates with an area of 1.75 cm2 and a separation distance of 4.33 mm. Suppose that the charge present on the two plates is equal to 2.67 picocoulombs.

Answer 1

The potential difference

$$V=\frac{q}{C}=\frac{qd}{\epsilon_0A}$$

$$V=\frac{2.67*10^{-12}*4.33*10^{-3}}{8.85*10^{-12}*1.75*10^{-4}}=7.46\:\rm V$$