Chapter 4: Q40P (page 1075)

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

Expert verified

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 $E_{\max}$ magnetic field $B_{\max}$ and it is given by equation (32.29) in the form $l = \frac{1}{2} ε_{0} {cE}_{\max}^{2}$ Also, The intensity I is proportional to the incident power P by area A and is given by $l = \frac{P}{A}$ 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 $l = \frac{P}{A}$ and $l = \frac{1}{2} ε_{0} c E_{m a x}^{2}$ we have,

$\frac{1}{2} ε_{0} {cE}_{\max}^{2} = \frac{P}{{πr}^{2}} \frac{E_{1}}{E_{2}} = \frac{r_{2}}{r_{1}} E_{2} = \frac{r_{1}}{r_{2}} E_{1} = \frac{10 m}{0.20 m} (3.50 N / C) = 175 N / C$