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

# Explanation: ## Step1: Recall energy - capacitor formula The energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}$, where $U$ is the energy, $C$ is the capacitance and $V$ is the voltage. We need to solve for $V$. Rearranging the formula for $V$, we get $V = \sqrt{\frac{2U}{C}}$. ## Step2: Substitute given values Given $C = 3.33\times10^{-6}\text{ F}$ and $U=8.15\times 10^{-4}\text{ J}$. Substitute these values into the formula: \[V=\sqrt{\frac{2\times(8.15\times 10^{-4}\text{ J})}{3.33\times10^{-6}\text{ F}}}\] First, calculate the value inside the square - root: $\frac{2\times8.15\times 10^{-4}}{3.33\times10^{-6}}=\frac{16.3\times 10^{-4}}{3.33\times10^{-6}}$. Using the rule of exponents $\frac{a\times10^{m}}{b\times10^{n}}=\frac{a}{b}\times10^{m - n}$, we have $\frac{16.3}{3.33}\times10^{-4+6}\approx4.895\times10^{2}$. Then, $V=\sqrt{4.895\times 10^{2}}\text{ V}\approx22.1\text{ V}$. # Answer: $22.1$