An air-filled capacitor has a potential difference between the plates of $80 \mathrm {~V}$. If the charge on each of the plates of the capacitor has magnitude $8.0 \mu \mathrm { C }$, what is the electrical energy stored by this capacitor?

Question

Quizplus · Accepted Answer

The electrical energy ($U$) stored in a capacitor can be calculated using the formula $U = \frac{1}{2} QV$, where $Q$ is the charge and $V$ is the potential difference. Substituting $Q = 8.0 \mu \mathrm{C} = 8.0 \times 10^{-6} \mathrm{C}$ and $V = 80 \mathrm{V}$, we get $U = \frac{1}{2} \times 8.0 \times 10^{-6} \mathrm{C} \times 80 \mathrm{V} = 320 \times 10^{-6} \mathrm{J} = 320 \mu \mathrm{J}$.

An Air-Filled Capacitor Has a Potential Difference Between the Plates 80 V80 \mathrm {~V}80 V

An Air-Filled Capacitor Has a Potential Difference Between the Plates $80 \mathrm {~V}$