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Comparison of Net Electric Force on a +q

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data
    In cases A and B shown there are two positive charges +Q each a distance s away from a third positive point charge +q. Is the net electric force on the +q charge in case A greater than, less than, or equal to the net electric force on the +q charge in case B? Explain.

    Here is a link to the stated question (question #2) with pictures showing Case A and Case B. http://www.csu.edu/chemistryandphysics/msabella/ccli/UpdatedLabs/physics2-calc/HW%20packet.pdf [Broken]




    2. Relevant equations

    I'm guessing using F=kqQ/r^2

    3. The attempt at a solution

    I'm not really sure where to begin, obviously there are two different angles in each case, but I'm not sure how to relate that to the force on the test charge?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 5, 2012 #2

    SammyS

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    Force is a vector quantity.

    Coulomb's Law, F=kqQ/r2, gives the magnitude of the force on q due to each charge Q.

    Do you know how to add vectors?
     
    Last edited by a moderator: May 5, 2017
  4. Mar 5, 2012 #3
    Yes, I do. However, I'm just confused because there is no real data here as far as angles are concerned. All we know is that Case B has a smaller angle than Case A. Just thinking about it, I thought that the smaller angle would mean a greater force, because the the two source charges are closer together and might mean a greater magnitude of force then Case A where the two source charges are farther apart?

    Meaning the distance between +q and +Q are the same, but the distance between both source charges are different in each case.
     
    Last edited: Mar 5, 2012
  5. Mar 5, 2012 #4

    SammyS

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    So, you have two cases: In each you're adding two vectors of equal magnitude. The only difference in the two cases is that in case A, the angle between the vectors is greater than the angle between the vectors in case B.

    Suppose that in case A, the vectors make an angle of θA with the horizontal. Similarly, in case B, the vectors make an angle of θB with the horizontal.

    Furthermore, let's suppose that θA = θB + Δθ , where θA, θB, and Δθ are all positive and all less than 90° .

    Can you find the magnitude of the resulting force in each case?
     
  6. Mar 6, 2012 #5
    So when adding the components would it be:

    Fnetx=kqQ/r^2 cos angle

    Fnety=kqQ/r^2 sin angle

    And since the angle is bigger in Case A, we would get a larger magnitude of the net force?
     
  7. Mar 6, 2012 #6

    SammyS

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    The y-components cancel, in each case.

    For angles between 0 and 90°, the cosine decreases as the angle increases.
     
  8. Mar 6, 2012 #7
    Ok, so the force would be greater in Case B then, because as the angle deceases cosine would be larger, making the magnitude greater?
     
  9. Mar 6, 2012 #8

    SammyS

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    Right !
     
  10. Mar 8, 2012 #9
    Great, Thanks!
     
  11. Mar 14, 2012 #10
    Now I have a similar question: two positive point charges +2Q and the effect on +q.

    Question: http://www.csu.edu/chemistryandphysics/msabella/ccli/UpdatedLabs/physics2-calc/HW%20packet.pdf [Broken]

    (2nd part of question #2 dealing with Case C and Case D)

    I'm just not sure where to start with this question as I'm not sure if there is a difference between Case C and D. Case C has two +2Q and Case D has four +Q. Would there be a difference, or are they equal in magnitude?
     
    Last edited by a moderator: May 5, 2017
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