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Coulombs Law - Finding Force When Altering Distance

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    If two charged bodies attract each other with a force of 1 newton, with what force will they attract each other if the distance between them is reduced to one-half of its original size? (The “newton,” abbreviated by the letter N, is the unit of force in the metric system. This is like the “pound” in the system that you are used to. Contrary to the way that most people use it, the kilogram is not a unit of force but a unit of mass.)

    2. Relevant equations

    F = k Qq/d2
    k = 9 x 10^9

    3. The attempt at a solution

    I sketched out my picture then a subsequent sketch with my distance cut in half:
    (+)---->1N 1N<----(-)
    (+)--> <--(-)

    I don't know how I am supposed to figure this new force without knowing the Q values or distance! I am only guessing that since the objects are twice as close, this causes the force to double. But honestly, I don't know how to back this up numerically without additional information! Someone please tutor me to understand these concepts better.
  2. jcsd
  3. Jan 21, 2012 #2


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    The force is inversely proportional to the square of the separation distance.
  4. Jan 21, 2012 #3
    I still don't understand what this means. So I take d^2 (even though I don't know what d is) and then take its inverse which means d^2= square root of d^2/1?? The thing I have always hated about physics and my brain is I don't see the connections that everyone else sees... How do you know what to do when I am only given 1 value? I don't even understand what is represented by the constant k. Textbooks and handouts are spread all over my desk and I am having such a hard time understanding where all these numbers are coming from. I am really in need of some elementary explanations. I am not trying to be lazy, I have been looking at this question for 2 hours now trying to figure out how to make sense of it.
  5. Jan 21, 2012 #4


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    The subject of proportions and inverse proportions, etc. is usually covered in fairly basic algebra courses. This topic essentially deals with ratios.

    The ratio of the new distance to the original is 1 : 1, a.k.a. 1/2

    The inverse of this is  ? 

    The square of that is  ? 
  6. Jan 21, 2012 #5

    The ratio of the new distance is 1:1? How is that ratio equal to 1/2?

    When I plug this into my equation, how to I solve from:

    1N=k(Qq/d^2(1/2)) I know that as my distance gets cut in half, my 1N will expand because of the inverse relationship.
  7. Jan 21, 2012 #6
    How am I expected to find the new force without a given distance??
  8. Jan 21, 2012 #7
    After watching about 20 youtube videos, I think I need to take 1/(1/2)^2 and I get 4N as my new force.

    I still don't understand how exactly this fits into the equation, but I am going to accept this answer with my lack of conceptual reasoning.

    Can someone explain to me what would happen if I were to cut one of my charges (q) in half?

    F=k (Qq(1/2))/d^2
  9. Jan 21, 2012 #8


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    When you're given an equation of several variables and constants (here you have F, k, Q, q, and d), and you're asked to determine how one particular variable is affected by a change in another variable, the usual approach is to take the given equation and form a ratio of the new value of the given variable to its original value. All this is done symbolically without needing to plug in actual values.

    In your case you have the formula [itex] F = k \frac{Q q}{d^2} [/itex]. You're told that d is being varied in some particular way and you want to know what happens to F. So create the ratio:

    [tex] \frac{F_{new}}{F_{old}} = \frac{k \frac{Q q}{{d_{new}}^2} }{k \frac{Q q}{{d_{old}}^2} } [/tex]
    Simplify the right hand side by cancelling common terms and that leaves you with:
    [tex] \frac{F_{new}}{F_{old}} = \frac{{{d_{old}}^2} }{{{d_{new}}^2} } = \left(\frac{{{d_{old}}} }{{{d_{new}}}}\right)^2 [/tex]
    Now, the problem says that the new distance is 1/2 the old distance, and that the original force is 1N. So
    [tex] \frac{F_{new}}{1N} = \left( \frac{d_{old}}{(1/2)d_{old}} \right)^2 [/tex]
    Simplify again and you're left with:
    [tex] F_{new} = 4N [/tex]
  10. Jan 21, 2012 #9
    ^ Very helpful. Thanks!
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