Electric field varying with distance

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Homework Help Overview

The discussion revolves around understanding the relationship between the amplitude of the electric field and distance, particularly in the context of electromagnetic waves and dipole antennas.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster seeks a simple explanation of the inverse relationship between electric field amplitude and distance. Some participants provide insights into the radiation pattern of dipole antennas and the mathematical relationship involving irradiance and electric field strength.

Discussion Status

Participants are exploring the concept of how electric field strength decreases with distance, referencing the relationship between irradiance and electric field. There is an ongoing exchange of ideas, but no consensus has been reached yet.

Contextual Notes

The original poster notes that only basic concepts of electromagnetic waves are covered in their textbook, indicating potential gaps in foundational knowledge that may affect understanding.

Jahnavi
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[Moderator note: moved from general physics, no template.]

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This is a solved example given in the book . Could someone help me understand how amplitude of electric field has an inverse relationship with distance ?

Only the very basics of EM waves are covered in the book so I would appreciate if someone could explain in a simple language .

Thank you .
 

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For a dipole antenna, the electric field of the radiated wave will stay parallel to the antenna dipole. This means the radiation pattern will mostly be 2 dimensional as a circle, centered at the transmitting antenna. As the circumference increases the field strength is "spread out" over more circumference.
So the electric field strength is divided by the length of the circumference. The circumference of a circle is proportional to r.
 
Recall that the irradiance (also called intensity) falls off as 1/r^2 (just think of the flux through spherical surfaces centered on the source) and is also proportional to E^2 (shown in just about every book, even if not derived). Putting both together,

I \propto 1/r^2 \propto E^2 => E \propto 1/r
 
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RedDelicious said:
Recall that the irradiance (also called intensity) falls off as 1/r^2 (just think of the flux through spherical surfaces centered on the source) and is also proportional to E^2 (shown in just about every book, even if not derived). Putting both together,

I \propto 1/r^2 \propto E^2 => E \propto 1/r

Thanks !
 

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