1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sunlight, Intensity, electric field RMS

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Estimate the rms electric field in the sunlight that hits Uranus, knowing that the Earth receives about 1350 W/m2 and that Uranus is 19.2 times farther away from the Sun (on average) than is the Earth.

    2. Relevant equations
    I=cεoErms2
    E=V/d

    3. The attempt at a solution
    Eearth=V/d
    Since Uranus is 19.2 times farther away then:
    Euranus=Eearth/19.2

    Iearth = 1350 W/m2
    Iearth=cεoEearth rms2
    Euranus rms = Eearth rms / 19.2 = √ ( I / (cεo)) / 19.2
    = 37.14 V/m

    I got this problem wrong several times so I just wanted to verify that I did it right this time. Thanks!
     
  2. jcsd
  3. Dec 4, 2016 #2
    Be aware that saying the electric field is equal to the electric potential divided by distance implies that you are assuming a constant electric field. In actuality the electric field equals the negative gradient of the potential.
     
  4. Dec 4, 2016 #3
    But yes, that answer does look right. I just calculated it using the relation between power and intensity, calculating the intensity for a distance 19.2 times greater, and plugging into your equation, and I came out with 37.15 V/m.
     
  5. Dec 4, 2016 #4
    Thanks! I got it right :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Sunlight, Intensity, electric field RMS
  1. Sunlight Intensity (Replies: 3)

Loading...