1. Limited time only! Sign up for a free 30min personal 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!

Energy associated with electric feild.

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    a conducting sphere of radius R has charge +Q on its surface. if the charge on the sphere is doubled and radius is halved, the energy associated with the electric feild will be?



    2. Relevant equations

    electric feild = K * Q/(R*R)

    Energy associated with feild will be (1/2)Integral of (ebsilon)E*E


    3. The attempt at a solution
    what am getting as answer is 64.but that is wrong. how can i solve this?
     
  2. jcsd
  3. Sep 23, 2011 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

  4. Sep 27, 2011 #3
    the answer should be 8 times. E = q/4pi*ebsilon*r*r. if we take it as E1 and E2 in the two cases and then compare them, we get the result. but my doubt is, in the question they have asked to find the energy associated with the electromagnetic field which i think is equal to integral(1/2)ebsilon*E*E. but that won't give the answer. the formula that gives the correct result is actually for finding electric feild at a point, rite? how can these two be the same?
     
  5. Sep 27, 2011 #4

    G01

    User Avatar
    Homework Helper
    Gold Member

    The answer is not 8.

    If the radius is halved and the charge is doubled, the electric field is increased by a factor of 4, not 64.

    Like you said: The energy depends on the square of the electric field. Thus, if the electric field increases by a factor of 4, by how much does the square of the electric field (and thus the energy) increase by?
     
  6. Sep 27, 2011 #5
    but sir, how can it be 4? as Q is double a factor 2 appears on the numerator. again as there is an (r/2) square on the denominator a factor 4 also appears in the numerator. so field in the 2nd case is 4*2=8 times the first one, rite?
     
  7. Sep 27, 2011 #6

    G01

    User Avatar
    Homework Helper
    Gold Member

    Oops! I forgot to square the 1/2! :redface: These little mistakes happen. My apologies for the confusion!

    Your answer of 64 is correct. The field is increased by a factor of 8, which means the energy is increased by a factor of 64.
     
  8. Sep 28, 2011 #7
    but sir, the thing is that options given does not include such an answer. given options are
    1. increases 4 times.
    2.increases 8 times
    3.remains the same
    4.decreases 4 time

    the answer should be 'increases 8 times'
     
  9. Sep 28, 2011 #8

    G01

    User Avatar
    Homework Helper
    Gold Member

    OK Assuming I am not misreading the question. Your answer of 64 is correct, regardless of what the book says. (It may be a typo.)

    We agree that the field is increased by a factor of 8:

    [tex]E\propto\frac{Q}{r^2}[/tex]

    so, [tex]E' \propto \frac{2Q}{(1/2)^2r^2}=\frac{8Q}{r^2}=8E[/tex]

    Therefore the energy:

    [tex]U\propto E^2[/tex]

    and [tex]U'\propto E'^2=8^2 E^2 = 64 U[/tex]
     
  10. Sep 28, 2011 #9
    Looks like a typo of some sort to me. Maybe they meant to ask how much the electric field increases by.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Energy associated with electric feild.
  1. Electric feild sketch (Replies: 2)

Loading...