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Homework Help: Please help me understand the inverse square law

  1. Jun 19, 2016 #1
    I am faced with the following question:
    Two point charges X and Y, exert a force F on each other when they are at a distance d apart (x and y are opposite charges). When the distance between them is 20mm, the force exerted on each other is 0.5F. What is the distance d?

    I know that, e.g. doubling r in F proportional to 1/r^2 will make F be 1/4 of the original, and tripling will make F 1/9 of the original but I do not know how it works the other way at all.

    It looks like the relationship would be r proportional to square root of 1/F... I do not understand this relationship.

    Could someone explain this in a very simple and understand way? I would also prefer this done as mathematically as possible....

    I am so troubled please I have an exam tomorrow and I just cannot understand how to get F from d value for 0.5F...
  2. jcsd
  3. Jun 19, 2016 #2


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    Seems like a straightforward application of Coulomb's Law:


    Note: Coulomb's Law and Newton's Law of Gravitation are both examples of Inverse Square Laws.
  4. Jun 19, 2016 #3
    I know the coulmob's law. I do not know Q1 or Q2 to solve it using coulomb is law. This is not helping. Please I need real help I just don't understand how to apply at all to these questions. I don't know if I am thick but I simply do not see it.. I have been crying over this question for over an hour please...

    Just an example of what if F was 0.21F when d=11, what would be d when 1F?
    Last edited: Jun 19, 2016
  5. Jun 19, 2016 #4
    You would set up an equation by representing the physical quantities (force and distance) with algebraic variables. When you plug in the numbers given in the problem you can solve for the unknown using elementary algebra.
  6. Jun 19, 2016 #5
    0.5F = k(Q1Q2/20^2)
    200F = K(Q1Q2)

    What can I do from here..? Am I doing the right thing even? I see 3 unknowns here...
  7. Jun 19, 2016 #6
    Please help...
  8. Jun 19, 2016 #7


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    You're in a panic and not understanding what information the problem is giving you.

    The problem says x and y are opposite charges, so what does that tell you about the magnitude of Q1 relative to Q2?

    When Q1 and Q2 are 20 mm apart, the force exerted on Q1 by Q2 (or vice versa) has magnitude 0.5 F.

    What does the distance d have to be between Q1 and Q2 so that the magnitude of the force exerted by one charge on the other is 1.0 F?

    This is a simple problem. All you need to do is calm down and set up a ratio between the two situations.

    If you don't get ahold of yourself soon, your exam will not go well.
  9. Jun 19, 2016 #8
    I am in a panic but I would not be able to solve this question if I weren't in a panic anyway.... My exam won't go down well anyway... I don't understand anything about the inverse square law or working with ratios in physics...

    I know I need 1F, and that it is somehow that d is 20x0.7 since d=14mm... But I don't get any eureka moments... I know that I need to double the F that is given at 20mm to get 1F... I do not understand what happens to r in F (proportional to) 1/r^2 when F gets double, I can only know that it would 1/4 if r got doubled... or 1/9 if it got tripled...

    Talking to me is like talking to a 3 year old child isn't it...
  10. Jun 19, 2016 #9
    I know I just said I have no eureka moments but... would this be the way to do ratios?

    0.5F : 1/20^2
    0.5 : 1/400
    1: 1/200
    1:1/10xSqrt of 2
    1: 1/14.14

    Although I don't know why it would be that you have to take the square root of 200...?
  11. Jun 19, 2016 #10


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    If you arrange your calculations in a systematic way, this goes a long way to provide clarity.

    The basic Coulomb's Law is:

    ##F = k \frac{Q_1 ⋅ Q_2}{r^2}##

    We are given that Q1 is the opposite charge of Q2, so that means that Q2 = -Q1.

    When d = 20 mm F = 0.5 ⋅ F, so we can substitute this information into Coulomb's Law equation:

    ##0.5 ⋅ F = k \frac {Q_1 ⋅ (-Q_1)}{0.020^2} = -k \frac{Q_1^2}{0.020^2}##

    ##F = -2 ⋅ k \frac{Q_1^2}{0.020^2}##

    Now, the alternate condition is to find d such that the force between charges is F:

    ##F = -k \frac{Q_1^2}{d^2}##

    Set the two expressions for F equal to one another and solve for the value of d:

    ##-k \frac{Q_1^2}{d^2} = -2 ⋅ k \frac{Q_1^2}{0.020^2}##
  12. Jun 19, 2016 #11
    In this problem, F is a constant, whereas in Coulomb's Law, F represents a variable.
  13. Jun 20, 2016 #12


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    And also we don't really need for ##Q_2=-Q_1##, the problem can be solved in a very similar way for any two charges , because the product ##Q_1Q_2## will appear in both sides of the equation and thus can be simplified.
  14. Jun 20, 2016 #13


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    No, but one needs to understand how to interpret the language of the problem statement, specifically "x and y are opposite charges".

    In many cases, too many IMO, a problem is not solved correctly not because the student is deficient in math or science, but because he or she cannot read and understand the problem properly.
  15. Jun 20, 2016 #14
    Good catch, Delta2. Since k, Q1 and Q2 are constants:

    F1(d1)2 = F2(d2)2

    1.0(d)2 = 0.5(20)2

    (Note that d is denominated in millimetres.)
  16. Jun 20, 2016 #15


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    What is the ratio of d to 20mm. (assuming d is also in units of mm) ?

    What is the ratio of F to 0.5F ?

    What is the relationship these two ratios?
  17. Jun 20, 2016 #16
    F1(d1)2 = F2(d2)2

    0.21(11)2 = 1.0(d)2

    (Note that in this problem, the values of Q1, Q2 or k are different than in the original problem you posted.)
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