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A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At \(10.0 \mathrm{~m}\) from this source, the amplitude of the electric field is measured to be \(1.50 \mathrm{~N} / \mathrm{C}\). What is the electric-field amplitude at a distance of \(20.0 \mathrm{~cm}\) from the source?

Short Answer

Expert verified
The electric-field amplitude at 20.0 cm is 75 N/C.

Step by step solution

01

Understanding the Inverse Square Law

The intensity of a wave spreading out uniformly from a point source decreases with the square of the distance from the source. This is known as the Inverse Square Law. Mathematically, the relation can be expressed as \( I \propto \frac{1}{r^2} \), where \( I \) is the intensity and \( r \) is the distance from the source.
02

Relating Intensity and Electric Field Amplitude

For electromagnetic waves, the intensity \( I \) is proportional to the square of the amplitude of the electric field \( E \). Thus, we have \( I \propto E^2 \). By combining these relations, we can relate the changes in the electric field amplitudes at different distances: \( \frac{E_1^2}{E_2^2} = \frac{r_2^2}{r_1^2} \).
03

Setting Up the Proportion

Given that \( E_1 = 1.50 \mathrm{~N/C} \) at \( r_1 = 10.0 \mathrm{~m} \). We are tasked with finding \( E_2 \) at \( r_2 = 0.20 \mathrm{~m} \). Substitute these values into the formula: \( \frac{1.50^2}{E_2^2} = \frac{0.20^2}{10.0^2} \).
04

Solving for the New Electric Field Amplitude

First, calculate \( 1.50^2 = 2.25 \) and \( (0.20/10.0)^2 = (0.02)^2 = 0.0004 \). Substitute and rearrange to solve for \( E_2^2 \): \( E_2^2 = 2.25 / 0.0004 \).
05

Calculating the Final Result

Solve for \( E_2 \): \( E_2^2 = 2.25 / 0.0004 = 5625 \). Then take the square root to find the electric field amplitude: \( E_2 = \sqrt{5625} = 75 \mathrm{~N/C} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electromagnetic Waves
Electromagnetic waves are fundamental to understanding how energy travels through space. These waves consist of oscillating electric and magnetic fields, which propagate perpendicular to each other and the direction of wave travel. Unlike mechanical waves, electromagnetic waves do not require a medium, enabling them to move through the vacuum of space.
Common examples include visible light, radio waves, and X-rays, all of which form part of the electromagnetic spectrum. Each electromagnetic wave travels at the speed of light, approximately \(3 \times 10^8 \text{ m/s}\), in a vacuum.

One essential aspect of electromagnetic waves is their ability to carry energy, quantified by the wave's intensity and related to the electric and magnetic field strengths. Understanding how the amplitude of the electric field varies with distance from the source is crucial for many practical applications, such as predicting signal strength in communication technologies.
Electric Field Amplitude
The electric field amplitude is a measure of how strong the electric component of an electromagnetic wave is at a particular location. It directly impacts the intensity of the wave, which indicates the energy flowing through a unit area per unit time.
Electric field amplitude is expressed in terms such as \(\text{N/C}\), which stands for newtons per coulomb, indicating the force experienced per charge due to the electric field.

The amplitude will change as the wave moves away from the source due to several factors. Applying the inverse square law, the amplitude decreases as the square of the distance increases. This phenomenon explains why signals from a radio tower diminish with distance, necessitating repeaters to maintain contact over long distances.
  • This relationship is vital in optimizing communication systems.
  • It helps engineers design more efficient signal transmitting and receiving systems.
  • Understanding amplitude changes helps in predicting power loss over distance.
By calculating the changing electric field amplitude, we can better manage and improve wireless communication systems.
Intensity and Distance Relationship
The intensity of an electromagnetic wave is intricately connected to the distance from its source, hence the focus on the intensity and distance relationship.

According to the inverse square law, intensity diminishes with the square of the distance from a point source. Mathematically, it can be expressed as \(I \propto \frac{1}{r^2}\), where \(I\) represents intensity and \(r\) denotes the distance.
This inverse relationship is due to the spreading out of wave energy over a larger area as the wave propagates. As the distance doubles, the energy spreads over four times the original area, effectively reducing the intensity.
  • In practical terms, understanding this concept allows for better placement of antennas and signal sources.
  • It also helps in calculating power levels needed for specific distances.
This knowledge is critical for fields such as wireless communication, radio broadcasting, and medical imaging, where energy absorption must be precisely controlled.

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Most popular questions from this chapter

If the eye receives an average intensity greater than \(1.0 \times 10^{2} \mathrm{~W} / \mathrm{m}^{2}\), damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in \(\mathrm{mW}\) ) that a laser beam \(1.5 \mathrm{~mm}\) in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in \(\mathrm{W} / \mathrm{cm}^{2} .\)

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{~Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from WCCO is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

In the 25 -ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity \(2500 \mathrm{~W} / \mathrm{m}^{2}\) at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor.

A sinusoidal electromagnetic wave having a magnetic field of amplitude \(1.25 \mu \mathrm{T}\) and a wavelength of \(432 \mathrm{~nm}\) is traveling in the \(+x\)-direction through empty space. (a) What is the frequency of this wave? (b) What is the amplitude of the associated electric field? (c) Write the equations for the electric and magnetic fields as functions of \(x\) and \(t\) in the form of Eqs. (12.17).

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