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The force when the distance is reduced

  1. May 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to one-ninth of its original value?

    F=1.5 N

    2. Relevant equations

    F=k|q1||q2|/r^2

    3. The attempt at a solution

    1.5 N = k |q1||q2|/(1/9)r^2

    I don't think I am setting this up right, but would I multiply 1.5N by 1/9 and by the constant k?

    Thanks in advance!
     
  2. jcsd
  3. May 30, 2013 #2

    Doc Al

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    Staff: Mentor

    Set it up so that 1.5 N is the force when the distance = R.

    Then see what the force is when the distance = R/9.

    (Set up a ratio.)
     
  4. May 30, 2013 #3

    Doc Al

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    Staff: Mentor

    Another way to approach this is to just ask yourself what happens to the force if the distance doubles? If the distance is cut in half? That should point you in the right direction.
     
  5. May 30, 2013 #4
  6. May 30, 2013 #5

    Doc Al

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    Try answering the questions I asked in post #3.
     
  7. May 30, 2013 #6
    If the distance doubles, the force increases x4 (exponentially?). If the distance is halved, the force decreases.
     
  8. May 30, 2013 #7

    Doc Al

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    Careful. The force is inversely proportional to the distance squared. So if the distance doubles, the force decreases by a factor of 4. (Multiplied by (1/2)2 = 1/4.)

    No, the force would increase by a factor of 4.

    F1 = (const)*(1/R)2

    If you reduce the distance by 2, replace R with R/2:

    F2 = (const)*(1/(R/2))2 = 4*(const)*(1/R)2 = 4*F1

    (Now do the same thing for the distance being reduced by 1/9.)
     
  9. May 30, 2013 #8
    F2 = (const)*(1/(R/9))2 = 81*(const)*(1/R)2 = 81*1.5 N = 121.5 N

    Wow, I really need to work on more of these.

    Thanks so much for your help!
     
  10. Apr 30, 2016 #9
    Hi Doc Al,

    I am trying to understand the answer that you posted above.

    F2 = (const)*(1/(R/2))2 = 4*(const)*(1/R)2 = 4*F1

    Could you explain why the squared sign is outside the last bracket. Should it not be inside the bracket as it is the distance squared.

    F2 = (const)*(1/(R/2)2)

    Also, how did you simplify it to be 4*(const), how did you calculate the 4?

    Many thanks,
    Cathal
     
  11. Apr 30, 2016 #10

    Doc Al

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    Realize that ##(\frac{1}{R})^2 = \frac{1}{R^2}##

    ##(\frac{1}{R/2})^2 = (\frac{2}{R})^2 = 4(\frac{1}{R})^2##
     
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