Question
Fernando detects the electric field from an isotropic source that is $22 \mathrm{km}$ away by tuning in an electric field with an rms amplitude of $55 \mathrm{mV} / \mathrm{m} .$ What is the average power of the source?
Step 1
The formula for intensity is given by: \[I = \frac{1}{2} \epsilon_0 E^2 c\] where \(I\) is the intensity, \(\epsilon_0\) is the permittivity of free space, \(E\) is the electric field, and \(c\) is the speed of light. Show more…
Show all steps
Your feedback will help us improve your experience
Mayukh Banik and 98 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
(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?
ELECTROMAGNETIC WAVES
Energy in EM Waves
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?
The $\mathrm{rms}$ value of the electric field in a certain light wave is $0.200 \mathrm{~V} / \mathrm{m}$. What is the amplitude of the associated magnetic field?
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD