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5 Charges

  1. Aug 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Four charges, Aq, Bq, Cq, and Dq (where q = 3.50 ×10−7 C), sit in a plane at the corners of a square whose sides have length d = 23.5 cm, as shown in the diagram below. A charge, Eq, is placed at the origin at the center of the square. (Note that A, B, C, D, and E are integer multipliers.)

    DATA: A = 2, B = 4, C = 5, D = 8, E = 1. Consider the charge at the center of the square, Eq.
    What is the net y-component of the force on this charge?

    Consider the situation where A=B=C=D=1, E=-1, and the following statements. Select "True" or "False" for each statement.
    1) The sum of the forces on the center charge in the y-direction equals zero.
    2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium.
    3)The equilibrium point at the center is a stable equilibrium for the motion of the negative charge in the plane of the square.
    4)The sum of the forces on the center charge in the x-direction does not equal zero.
    5)If one were to double the magnitude of the upper-right-hand positive charge, the negative charge would be in equilibrium.
    6)The equilibrium point at the center is an unstable equilibrium for the motion of the negative charge in the line from the center charge and perpendicular to the plane of the square.


    2. Relevant equations
    [tex] F= \frac{Kq_{1}q_{2}}{r^2}[/tex]



    3. The attempt at a solution
    So for the first one I break up each each force acting on E. I then do [tex]Fsin(\theta)=F_{y}[/tex] for each one and take the sum right, i keep getting it wrong for some reason.
    For the second part:
    1)T
    2)T
    3)T
    4)F
    5)F
    6)F
    Thanks
     

    Attached Files:

    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2
    is it 0.197N?
     
  4. Aug 25, 2007 #3
    How did you get that? I got 0.1725
    What did you get fi the y components of each charge?
     
  5. Aug 25, 2007 #4
    So, first find E (net Electric field) at point E.

    what was your r distance? - it should be same for all points
    and write charge values like p(q) where p is the integer and q is the q value
    and sin(theta) for all is same = 1/sqrt(2)

    so, net E = sin(thea)*k*q*(1/r^2)[p[A]+p+p[C]+p[D]]
    factored out everything that was common.

    so try it.

    Edit: from net E I meant E in y direction
     
    Last edited: Aug 25, 2007
  6. Aug 25, 2007 #5
    so y didn't you use [tex] F= \frac{Kq_{1}q_{2}}{r^2}[/tex] on each pair of charges?
     
  7. Aug 25, 2007 #6
    you can also do that, but I preferred to find electric field first, it really doesn't matter much
     
  8. Aug 25, 2007 #7
    Oh ok. What do you think about the T/F questions?
     
  9. Aug 25, 2007 #8
    I didn't get most of them ><.

    like
    "2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium."

    I don't know where's the negative charge.

    And 1) shouldn't it be F, we just found a value for it?

    Edit: oops.. I missed that sentence..so wait for like a sec

    Edit3: I don't know #3 amd #6, but for others I also got the same answers.
     
    Last edited: Aug 25, 2007
  10. Aug 25, 2007 #9
    See that is where I am stuck also.
     
  11. Aug 25, 2007 #10
    So any ideas on the true/false?
     
  12. Aug 25, 2007 #11
    nopes
     
  13. Aug 25, 2007 #12
    anyone? pretty confident on the true false except for 3 &6. Aren't they both kind of asking the same thing though? The central charge E is stable in its plane and move out of it?
     
  14. Aug 25, 2007 #13

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    This is all for A=B=C=D=1 and E=-1

    1) The sum of the forces on the center charge in the y-direction equals zero. T

    2)If one were to triple the magnitude of the negative charge, the negative charge would be in equilibrium. T

    3)The equilibrium point at the center is a stable equilibrium for the motion of the negative charge in the plane of the square. T

    4)The sum of the forces on the center charge in the x-direction does not equal zero. F

    5)If one were to double the magnitude of the upper-right-hand positive charge, the negative charge would be in equilibrium. F

    6)The equilibrium point at the center is an unstable equilibrium for the motion of the negative charge in the line from the center charge and perpendicular to the plane of the square. F
     
  15. Aug 25, 2007 #14

    learningphysics

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    Hmmm... I'm getting
    7*sin(45)kq^2/r^2
    =7*0.7071068*9E9(3.5E-7)^2/(0.235/2)^2
    =0.395N for the first part...

    Oops... should be:

    7*sin(45)kq^2/r^2
    =7*0.7071068*9E9(3.5E-7)^2/[(0.235/2)^2]*2
    =0.1976N for the first part...
     
    Last edited: Aug 25, 2007
  16. Aug 25, 2007 #15
    yah first part is right but T & F is wrong.
     
  17. Aug 25, 2007 #16

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    Oops... yeah, 3 and 6 should be opposite... 3 is false. 6 is true. Because when the negative charge is moved from the center, it will be attracted towards one end...
     
  18. Aug 25, 2007 #17

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    0.395N is right?
     
  19. Aug 25, 2007 #18
    The answer to the first part is .197N. Second part is still wrong.
     
  20. Aug 25, 2007 #19

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    Oops... both 3 and 6 are false...

    The negative charge is in unstable equilibrium for motion within the plane... but it is in stable equilibrium for motion perpendicular to the plane...

    Reason is it will be attracted back downwards towards the plane, back to the equilibrium point...
     
  21. Aug 25, 2007 #20

    learningphysics

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    Yeah, I messed up... now I get 0.1976N...
     
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