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Chapter 22: Problem 59

What is the maximum power level of a radio station so as to avoid electrical breakdown of air at a distance of 0.65 m from the transmitting antenna? Assume the antenna is a point source. Air breaks down in an electric field of about \(3\times 10^6\) V/m.

Short Answer

Expert verified
Maximum power level is approximately \(5.23 \times 10^5\, \text{W}\).

Step by step solution

01

Understand the Problem

We need to find the maximum power level of a radio station with an antenna acting as a point source, given that the air breakdown electric field is \(3 \times 10^6\, \text{V/m}\) at a distance of 0.65 meters.
02

Relate Electric Field and Power

A point source antenna emits electromagnetic waves, and the maximum electric field (breakdown field) relates to the power \(P\) through the formula for the electric field from a point source: \(E = \sqrt{\frac{2P}{c \varepsilon_0}} \cdot \frac{1}{r}\), where \(c\) is the speed of light, \(\varepsilon_0\) is the permittivity of free space, and \(r\) is the distance from the source.
03

Substitute Known Values

Substitute the known values: \(E = 3 \times 10^6\, \text{V/m}\), \(r = 0.65\,\text{m}\), \(c = 3 \times 10^8\, \text{m/s}\), and \(\varepsilon_0 = 8.85 \times 10^{-12}\, \text{F/m}\) into the equation from Step 2: \[3 \times 10^6 = \sqrt{\frac{2P}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \cdot \frac{1}{0.65}\]
04

Solve for Power \(P\)

Solve the equation for \(P\). First, square both sides to eliminate the square root: \[(3 \times 10^6 \cdot 0.65)^2 = \frac{2P}{3 \times 10^8 \times 8.85 \times 10^{-12}}\]Simplify and solve for \(P\):\[P = \frac{(3 \times 10^6 \times 0.65)^2 \cdot 3 \times 10^8 \times 8.85 \times 10^{-12}}{2}\]
05

Calculate Numerical Answer

Calculate the numerical value:\[(3 \times 10^6 \times 0.65)^2 = 3.94725 \times 10^{12}\]\[P = \frac{3.94725 \times 10^{12} \times 3 \times 10^8 \times 8.85 \times 10^{-12}}{2} = 5.23 \times 10^5\, \text{W}\]
06

Conclude the Result

The maximum power level such that the radio waves do not cause electrical breakdown of air at a distance of 0.65 m is approximately \(5.23 \times 10^5\, \text{W}\).

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Key Concepts

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

Electric Field Strength
Electric field strength is a measure of the force experienced by a unit electric charge placed in a field. It is represented by the symbol \( E \) and measured in volts per meter (V/m).
In electrostatics, the electric field around a charge is defined as the force per unit charge that a charged object experiences.
For electromagnetic waves, like those emitted by a radio station, the electric field strength is crucial because it determines how energy is transmitted through the air.
  • When the electric field exceeds a certain level, known as the breakdown field, the air becomes ionized and can conduct electricity.
  • For air, the breakdown electric field is about \(3 \times 10^6\) V/m.
Understanding electric field strength also helps in designing safe systems that prevent electrical breakdown, ensuring that emitted electromagnetic waves do not exceed safe limits.
Power of Radio Station
A radio station's power refers to the amount of energy it emits as electromagnetic waves per second, measured in watts (W).
The power of a radio station determines the range and intensity of the transmitted signal.
When analyzing the relationship between power and electric field strength, it's essential to understand the impact of a point source antenna.
  • The electric field strength \( E \) at a distance \( r \) from such an antenna is given by the equation \( E = \sqrt{\frac{2P}{c \varepsilon_0}} \cdot \frac{1}{r} \).
  • This relationship shows that the electric field strength is directly related to the square root of the power \( P \) and inversely proportional to the distance from the antenna.
Therefore, understanding this relation is key to determining the maximum permissible power level of a radio station to avoid electrical breakdown.
Permittivity of Free Space
The permittivity of free space, denoted by \( \varepsilon_0 \), is a fundamental physical constant that characterizes how electric fields interact with vacuum.
It reflects the capacity of a vacuum to permit electric field lines and is measured in farads per meter (F/m).
The permittivity of free space is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
  • It plays a vital role in defining the strength of interactions between charges in vacuum.
  • In the context of electromagnetic waves from a radio station, \( \varepsilon_0 \) is part of the equation relating electric field strength \( E \) with power \( P \): \( E = \sqrt{\frac{2P}{c \varepsilon_0}} \cdot \frac{1}{r} \).
Knowing this constant helps predict how electromagnetic waves propagate in a vacuum, which is important in understanding their behavior in air as well.

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Most popular questions from this chapter

(II) At a given instant, a 3.8-A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side?

(I) If the electric field in an EM wave has a peak magnitude wave has a peak magnitude of \(0.72\times 10^-4\) V/m, what is the what is the peak magnitude of the magnetic field strength?

A global positioning system (GPS) functions by determining the travel times for EM waves from various satellites to a land-based GPS receiver. If the receiver is to detect a change in travel distance on the order of 3 m, what is the associated change in travel time (in ns) that must be measured?

(II) A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms value of the electric field?

(I) An EM wave has a wave length of 720 nm. What is it's frequency, and how would we classify it?
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