Question

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At a distance of 10.0 m from this source, the amplitude of the electric field is measured to be 3.50 N/What is the electric-field amplitude 20.0 cm from the source?

Short answer

At 20 cm, the amplitude of the electric field is 175 N/C

Step-by-step solution

Concept of the intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field

The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude ${\mathbf{E}}_{\mathbf{max}}$magnetic field ${\mathbf{B}}_{\mathbf{max}}$and it is given by equation (32.29) in the form $\mathbf{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{cE}}_{\mathbf{max}}^{\mathbf{2}}$Also, The intensity I is proportional to the incident power P by area A and is given by $\mathbf{l}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$Where r is the radius and represents the distance from the source.

Calculate the amplitude

At r1 = 10 m, the electric field amplitude is El = 3.50 N/c and we need to find E2 at r2 = 20 cm. From equation$\mathrm{l}=\frac{\mathrm{P}}{\mathrm{A}}$ and$l=\frac{1}{2}{\epsilon }_{0}c{E}_{max}^2$ we have,

$$\begin{gathered} \frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{cE}}_{\max }^2=\frac{\mathrm{P}}{{\mathrm{\pi r}}^2} \\ \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1} \\ {\mathrm{E}}_2=\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}{\mathrm{E}}_1 \\ =\frac{10\mathrm{m}}{0.20\mathrm{m}}\left(3.50\mathrm{N}/\mathrm{C}\right) \\ =175\mathrm{N}/\mathrm{C} \end{gathered}$$

Therefore, the amplitude is 175N/C