A sinusoidal electromagnetic wave has a peak electric field of $8.00 \mathrm { kV } / \mathrm { m }$ What is the intensity of the wave? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \right.$ $\mu _ { 0 } = 4 \pi \times 10 ^ { - 7 } \mathrm {~T} \cdot \mathrm { m } / \mathrm { A } ,$ $\left. \varepsilon _ { 0 } = 8.85 \times 10 ^ { - 12 } \mathrm { C } ^ { 2 } / \mathrm { N } \cdot \mathrm { m } ^ { 2 } \right)$ A)170 kW/m² B)85 kW/m² C)21 kW/m² D)11 kW/m²

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Quizplus · Accepted Answer

The intensity $I$ of an electromagnetic wave is given by the formula $I = \frac{1}{2} c \varepsilon_0 E^2$, where $c$ is the speed of light, $\varepsilon_0$ is the permittivity of free space, and $E$ is the peak electric field. Substituting the given values, $I = \frac{1}{2} \times 3.0 \times 10^8 \times 8.85 \times 10^{-12} \times (8.00 \times 10^3)^2 = 85 \times 10^3 \, \text{W/m}^2$ or $85 \, \text{kW/m}^2$.

A Sinusoidal Electromagnetic Wave Has a Peak Electric Field Of 8.00