Question

Two 0.585 cm x 0.585 cm square aluminum electrodes, spaced 0.5701 mm apart are connected to a 40 V battery. What is the charge on each plate in nanoCoulombs?

Answer 1

Given

The dimmension of the electrode is 0.585 cm x 0.585 cm

The distance is d=0.5701 mm

The potential is V=40 V

To find

What is the charge on each plate in nanoCoulombs?

Explanation

The capacitance is

$\begin{gathered} C=\frac{\epsilon_oA}{d} \\ \Rightarrow C=8.85\times10^{-12}\times\frac{0.585\times0.585\times10^{-4}m^2}{0.5701\times10^{-3}} \\ \Rightarrow C=5.31\times10^{-13}F \end{gathered}$

The charge on each plate is

$\begin{gathered} q=CV \\ \Rightarrow q=5.31\times10^{-13}\times40 \\ \Rightarrow q=2.12\times10^{-11} \\ \Rightarrow q=0.02nC \end{gathered}$

Conclusion

The charge is 0.02 nC.