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Studying for my SAT II Physics

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  1. May 31, 2008 #1
    i'm studying for my SAT II Physics on June 7, 2008. and taking practice tests from REA.
    i had trouble understanding this question. please help :) it would be VERY appreciated.

    1. The problem statement, all variables and given/known data
    Two charges are separated by 2 m. The force of attraction between them is 4 N. If the distance between them is doubled, the new force between them is...
    A) .5 N
    B) 1 N
    C) 2 N
    D) 4 N
    E) 8 N

    2. Relevant equations
    Force = [K(Q1)(Q2)] / (R^2)
    the formula is on http://www.sparknotes.com/testprep/books/sat2/physics/chapter13section2.rhtml under Coulomb's Law

    3. The attempt at a solution
    i'm sorry, i really couldn't figure out. it's probably a simple question, and i just can't see it.

    PLEASE HELP!
    thank you SO much :biggrin:
     
  2. jcsd
  3. May 31, 2008 #2
    Hello,
    You are given the distance = 2m. In Coulomb's Law the distance is R.
    Now, if the distance doubles, then will the force increase or decrease(i.e. will it be greater or equal to 4N)? and by what factor?
     
  4. May 31, 2008 #3
    Ummm....the force would decrease because the force is inversely proportional to the square of distance. And it would decrease by a factor of 1/16, right?
     
  5. Jun 1, 2008 #4

    alphysicist

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    Homework Helper

    Hi oceanflavored,

    It would decrease, and the new denominator is 16, but that's not the factor that the force decreased by. You have to take into account what the force was originally.
     
  6. Jun 1, 2008 #5
    eureka!! (i think!)

    okay, i think i get this:
    if the distance increases by a factor of 2, the force would decrease by a factor of (1/2)^2 or (1/4)
    so the new force would be 4 x (1/4) = 1 N
    yessssss???
     
  7. Jun 1, 2008 #6
    Yes, that's correct.
     
  8. Jun 1, 2008 #7

    tiny-tim

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    Hi oceanflavored! :smile:

    yessssss!!! :biggrin:
     
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