Electric field strength from a light bulb

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SUMMARY

The electric field strength of electromagnetic waves from a 100-W light bulb, measured at a distance of 3.0 meters, can be calculated using the relationship between electromagnetic flux and electric field amplitude. The relevant equations include the electric field equation, vec{E}=\frac{kq}{d^{2}}, and the power equation, P=\frac{2kq^{2}a^{2}}{3c^{3}}. A critical aspect of this calculation involves understanding the flux of electromagnetic energy through a sphere surrounding the light bulb, rather than assuming a monopole charge model.

PREREQUISITES
  • Understanding of electromagnetic wave theory
  • Familiarity with the equations for electric field strength and power
  • Knowledge of the constants involved, such as k (Coulomb's constant) and c (speed of light)
  • Basic calculus for derivatives and integrals
NEXT STEPS
  • Research the concept of electromagnetic flux and its relation to electric field strength
  • Study the derivation and application of the electric field equation vec{E}=\frac{kq}{d^{2}}
  • Learn about the characteristics of electromagnetic waves and their propagation
  • Explore the differences between monopole charge models and actual light bulb operation
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in calculating electric field strengths from light sources.

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Homework Statement


What is the approximate electric field strength of the electromagnetic waves radiated by a 100-W light bulb, measured 3.0m from the bulb?

Homework Equations


$$\vec{E}=\frac{kq}{d^{2}}$$

$$P=\frac{2kq^{2}a^{2}}{3c^{3}}$$

The Attempt at a Solution


I solved for ##q=\sqrt{\frac{3Pc^{3}}{2ka^{2}}}##, and then plugged that into the equation for electric field, and I have all the values I need to solve except for the acceleration, which I'm not sure how to find. I thought about using the second derivative of ##x(t)= x~sin(\omega~t)##, but I don't know what to use for ##x##, ##\omega## or ##t##.
 
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Your model assumes that lightbulb filaments produce light by having a monopole charge ##q## that radiates energy. That's not how light bulbs work. For this calculation, you need to consider the flux of electromagnetic energy through a sphere of radius 3.0 m and relate that flux to the amplitude of the electric field in the electromagnetic plane wave that crosses the sphere.

Edited to correct typo.
 
Last edited:
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