Question

A parallel-plate capacitor has 4.00cm² plates separated by 6.00mm of air. If a 12.0V battery is connected to this capacitor, how much energy does it store in Joules?

Answer 1

Since we have a parallel capacitor with a dielectric (air at room temperature, which has a dielectric constant K=1.00059), we use the following equation with the area of the plates (A = 4 cm²) and distance (d = 6 mm = 0.6 cm) between the plates to first calculate the capacitance:

$C=\frac{Q}{V}=K\epsilon_0\frac{A}{d}$

$C=(1.00059)(8.854187817\times10^{-12}\frac{C^2}{J\cancel{m}})(\frac{4\,\cancel{cm^2}}{0.6\,\cancel{cm}}*\frac{1\,\cancel{m}}{100\,\cancel{cm}})$

$C=5.9063\times10^{-13}\frac{C^2}{J}$

Then, since we know that V = 12 V = 12 J/C and that the energy that can be stored on any capacitor is U = $\frac{CV^2}{2}$, we substitute and find U:

$U=\frac{(5.9063\times10^{-13}\cancel{\frac{C^2}{J}})(12\,\frac{J}{\cancel{C}})^2}{2}=4.2525\times10^{-11}\,J$

In conclusion, the energy stored on this capacitor is

U = 4.2525 x 10^-11 J.

Reference:

• Young, H. D., & Freedman, R. A. (2014). Sears and Zemansky's University Physics (with Modern Physics Technology Update). Edinburgh: Pearson.