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Homework Help: F = kQq/R2 finding force problem

  1. Jul 30, 2011 #1
    Two charges Q1 amd Q2 jabe a force of 3.1x10^-8 N when the distance between them is 3m. Use the formula F = kQq/R2 to determine the force when the distance (R) between charge Q1 and Q2 increased from 3m to 4.5m

    3. The attempt at a solution

    My attempt was that
    F = 1/R^2
    k= 9.0x10^9


    3.1x10^-8 / 0.44
    =.7 x10^-8

    The answer being F = .7 x10^-8 N

    Is that correct?
  2. jcsd
  3. Jul 30, 2011 #2


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    Up to (3/4.5)^2 I was fairly comfortable with.

    From there on, not so much.
    I think 0.44N was not such a great statement, but I know what you meant [it reads like a force value in Newtons]

    Next I am not sure why you decided to divide by 0.44 - nor why you thought the answer was what you gave.

    I don't think you should round of 4/9 to 0.44, and think more carefully about how to use that factor.
  4. Jul 30, 2011 #3
    Yes good idea, i'll stick with 4/9 instead of the decimal to preserve accuracy. Since F is proportional to 1/R^2. If R increases from 3 m to 4.5 m shouldn't the force decrease by a factor of (3/4.5)^2. And i thought decrease by a factor means divided by so i divide the original force by the force i got from the divide i did prior?

    Edit: I just divided it the 4/9, it gives me 6.9x10^-8 ( i see why you suggested i'd use 4/9 )
  5. Jul 30, 2011 #4


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    It is common for people to confuse what to do with the factors.

    You calculated a factor of 4/9 [good], and wanted to find the reduced force. That is, you wanted to get a smaller value for the Force.

    How do you use 4/9 to get a smaller value?

    Note: Some people get a factor of 9/4 [perhaps in error]. what should they do with a factor of 9/4 in order to get a smaller value?

    And don't forget to apply your factor to the original value given!!
  6. Jul 30, 2011 #5
    Hmm so was my original divide procedure justified from my explanation?
    3.1x10^-8 / (4/9) = 6.9x10^-8 ...the force increased so this has to be wrong since it has to decrease right?
  7. Jul 30, 2011 #6


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    Yes - it is supposed to decrease.

    I never bother too much which factor my students calculate [4/9 or 9/4], I just want them to think about how to effect the change needed.
    If you want a smaller answer, and will be using a factor less than 1, you need to multiply.
    If you want a smaller answer, and will be using a factor greater than 1, you need to divide.

    Send me back your new answer then I will explain how I would have looked at/solved the problem.
  8. Jul 31, 2011 #7


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    Did you sort out where you went wrong? Only a single, simple error!
  9. Aug 1, 2011 #8
    Sorry for the late reply, had to work and family friends were visiting.

    Would you then have to multiply it since 4/9 is under 1.
    3.1x10^-8 x 4/9 = 1.3 x 10^-8

    Would the force be just 1.3x10^-8 or would i have to subtract that from 3.1x10^-8?
  10. Aug 1, 2011 #9


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    The multiplication is correct. Though you have merely "truncated" the answer rather than round it off - it should be 1.4 x 10^-8 when rounded to 2 figures.

    Fortunately if you had subtracted 4/9 form 3.1 x 10^8 you would get approximately -4/9 and thus recognise it as wrong.

    4/9 was a factor, and you either multiply or divide by factors.

    Fractions are always a problem, because often work backwards: multiplying by them you get smaller, divide you get bigger.

    The way I would have done it - which ultimately is numerically the same thing - is.

    Since separation is changing from 3 to 4.5 - not an integer factor change - I would work on an interim separation of 1.5 m

    This is half the first separation - meaning 4 x the force or 12.4 x 10^-8

    I then change to 4.5.

    This is 3 times the interim separation so it if F / 9 or 1.377778 x 10^-8

    Rounding to 2 significant figure this is 1.4 x 10^-8

    ********** Be careful to always do calculations to as many figures as your calculator can handle - then at the end ROUND OFF, don't just chop off the last figures.**************
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