a 3.33 x 10-6 f capacitor stores 8.15 x 10-4 ...

# Explanation: ## Step1: Recall energy - charge - capacitance formula The energy stored in a capacitor is given by $U=\frac{Q^{2}}{2C}$, where $U$ is the energy, $Q$ is the charge, and $C$ is the capacitance. We need to solve for $Q$. Rearranging the formula for $Q$, we get $Q = \sqrt{2UC}$. ## Step2: Substitute given values Given $C = 3.33\times10^{-6}\text{ F}$ and $U=8.15\times 10^{-4}\text{ J}$. $Q=\sqrt{2\times(8.15\times 10^{-4}\text{ J})\times(3.33\times10^{-6}\text{ F})}$ First, calculate the product inside the square - root: $2\times8.15\times 10^{-4}\times3.33\times10^{-6}=2\times8.15\times3.33\times10^{-4 - 6}=54.099\times10^{-10}$. $Q=\sqrt{54.099\times10^{-10}}$. We can rewrite $54.099\times10^{-10}$ as $5.4099\times10^{-9}$. $Q=\sqrt{5.4099\times10^{-9}}=\sqrt{5.4099}\times\sqrt{10^{-9}}\approx2.326\times10^{-4.5}\approx2.33\times 10^{-5}\text{ C}$ # Answer: $2.33\times 10^{-5}$