A sinusoidal electromagnetic wave has a peak electric field of $8.00 \mathrm { kV } / \mathrm { m }$. What is the intensity of the wave? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } , \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The intensity $I$ of an electromagnetic wave can be calculated using the formula $I = \frac{1}{2} c \varepsilon_0 E_0^2$, where $c$ is the speed of light, $\varepsilon_0$ is the permittivity of free space, and $E_0$ is the peak electric field. Substituting the given values, $I = \frac{1}{2} (3.0 \times 10^8 \, \text{m/s}) (8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2) (8.00 \times 10^3 \, \text{V/m})^2$, results in an intensity of approximately $85 \, \text{kW}/\text{m}^2$.

A Sinusoidal Electromagnetic Wave Has a Peak Electric Field Of 8.00