An electromagnetic wave has an electric field amplitude of $E_0=100\text{ V/m}$. What is the average energy density $u_{avg}$ of the wave in free space?

Solution

1. The average energy density $u_{avg}$ of an electromagnetic wave is given by the formula $u_{avg}=\frac{1}{4}{ϵ}_0{E}_0^2$, where ${ϵ}_0\approx 8.85\times 10^{-12}\text{ F/m}$ and $E_0=100\text{ V/m}$.
2. Substitute the given values into the equation: $u_{avg}=\frac{1}{4}(8.85\times 10^{-12}\text{ F/m})(100\text{ V/m}{)}^2$.
3. Simplify the expression: $$\begin{array}{rl}{u}_{avg}& =0.25\times 8.85\times 10^{-12}\times 10^4\\ & =2.2125\times 10^{-8}{\text{ J/m}}^3\end{array}$$.
4. Hence the answer is (A).