a 4.03 x 10^-5 f capacitor stores 5.15 x 10^-3 j of energy. what is the voltage across the capacitor? ? v

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 $V$.

Step2: Rearrange the formula for $V$

Starting from $U = \frac{1}{2}CV^{2}$, we can first multiply both sides by 2 to get $2U=CV^{2}$. Then divide both sides by $C$: $V^{2}=\frac{2U}{C}$. Taking the square - root of both sides gives $V=\sqrt{\frac{2U}{C}}$.

Step3: Substitute the given values

We are given that $U = 5.15\times10^{-3}\text{ J}$ and $C = 4.03\times10^{-5}\text{ F}$. Substituting these values into the formula $V=\sqrt{\frac{2\times(5.15\times10^{-3})}{4.03\times10^{-5}}}$. First, calculate the value inside the square - root: $\frac{2\times(5.15\times10^{-3})}{4.03\times10^{-5}}=\frac{10.3\times10^{-3}}{4.03\times10^{-5}}$. Using the rule of exponents $\frac{a\times10^{m}}{b\times10^{n}}=\frac{a}{b}\times10^{m - n}$, we have $\frac{10.3}{4.03}\times10^{-3+5}\approx2.56\times10^{2}$. Then, $V=\sqrt{2.56\times10^{2}} = 16\text{ V}$.

Answer:

16