What is the electric field amplitude 2m from the source

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SUMMARY

The discussion centers on calculating the electric field amplitude at a distance of 2 meters from a 60 W monochromatic point source in a vacuum. The intensity is calculated as 1.19 W/m² using the formula Intensity = Power / (4 * Pi * r²). The relationship between intensity and electric field amplitude is established through the equation I = (1/2)ε₀cE², where ε₀ is the permittivity of free space and c is the speed of light. Additionally, a secondary question regarding the logarithmic relationship between intensity and voltage across a light-dependent resistor (LDR) in a potential divider circuit is raised, specifically focusing on a linear relationship with a slope of -1.7.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically electric field concepts.
  • Familiarity with intensity calculations in physics.
  • Knowledge of logarithmic functions and their properties.
  • Basic principles of circuits, particularly potential divider circuits and LDRs.
NEXT STEPS
  • Study the derivation of the electric field amplitude from intensity using I = (1/2)ε₀cE².
  • Learn about the properties of logarithmic functions and their applications in physics.
  • Research the behavior of light-dependent resistors (LDRs) in potential divider circuits.
  • Explore the relationship between intensity and distance from a point source in more complex scenarios.
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the practical applications of electric field calculations and circuit analysis.

kingyof2thejring
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Homework Statement


A 60 W monochromatic point source is radiating equally in all diresctions in a vacum. What is the electric field amplitude 2m from the source.


Homework Equations


Intensity=Power/(4*Pi*r^2), Intensity proportional to square Amplitude


The Attempt at a Solution


I=1.19 Wm-2 what happens next?
 
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Use I=(1/2)\epsilonzero c E^2.
 
Desperate semi related question. If you have a graph of log intensity against log of (voltage across an LDR in a potential divider circuit)
how do you explain the relationship if your resultant is a straight line of -1.7 obviously in the form of y=mx+c
Log(y)= nlog(x) + logm where y is intensity and x is voltage n being -1.7 and logm being y intercept.
im trying to explain the relationship between the two quantities and ... I am stuck if anyone can help me i would very much appreciate it...
 
hopeless blonde said:
Desperate semi related question. If you have a graph of log intensity against log of (voltage across an LDR in a potential divider circuit)
how do you explain the relationship if your resultant is a straight line of -1.7 obviously in the form of y=mx+c
Log(y)= nlog(x) + logm where y is intensity and x is voltage n being -1.7 and logm being y intercept.
im trying to explain the relationship between the two quantities and ... I am stuck if anyone can help me i would very much appreciate it...

Welcome to PF, hopeless blonde. In future, however, please don't hijack another person's thread with your own question-- instead, start a new thread in the relevant homework section.

With regard to your question, you have an equation of the form logy=nlogx+logm. Now, nlogx can be expressed as log(xn). Do you know a rule for loga+logb?
 

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