Chapter 15: Problem 2171
A point source of electromagnetic radiation has an average output power of \(800 \mathrm{~W}\). The maximum value of electric field at a distance of \(4.0 \mathrm{~m}\) from the source is (A) \(64.7 \mathrm{Vm}^{-1}\) (B) \(57.8 \mathrm{Vm}^{-1}\) (C) \(56.72 \mathrm{Vm}^{-1}\) (D) \(54.77 \mathrm{Vm}^{-1}\)
Short Answer
Step by step solution
Write down the given variables
Find the intensity of radiation at the given distance
Calculate the intensity
Write down the formula for the Poynting vector
Calculate the maximum electric field
Choose the correct answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Field
Intensity of Radiation
- The formula used is \( I = \frac{P}{A} \), where \( P \) is the power and \( A \) is the area.
- For a point source radiating uniformly in all directions, \( A \) is the surface area of a sphere, calculated as \( 4 \pi r^2 \).
Poynting Vector
- \( \varepsilon_0 \) is the vacuum permittivity.
- \( c \) is the speed of light.
- \( E_{max} \) is the maximum electric field.
Vacuum Permittivity
Speed of Light
- It determines the maximum speed at which information or matter can travel.
- Within the Poynting vector formula, \( c \) is crucial for linking the electric field's maximum value to the intensity of the wave.
- This speed underpins many calculations in electromagnetic theory and is central to understanding how electromagnetic waves propagate over distances.