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Four charges each Q in magnitude

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Four charges, each q in magnitude, are arranged in a square, two positive and two negative, in opposite corners. The square is L length on each side, and the charges are all in the same plane as shown in the diagram. Show work and units for full credit.

    q+ -----B----- q-
    | |
    | | L
    | |
    q- ------------ q+
    L
    **vertical dotted line makes square but spacing corrects itself

    A. Determine the magnitude and direction of the total electric field at point B (half way between the top two charges) due to all four charges as a function of q and L.

    B. If placed at point B, which would accelerate faster, object 1 with a charge 2Q and mass M or object 2 with a charge Q and mass M/2? Why?

    C. Draw the electric field lines in the plane of the square indicating the direction and general shape of the electric field generated by this charge arrangement. Hint: think about the electric field generated by two opposite charges.

    D. Is the electric field zero at any point(s)? If so, where? Also, indicate any equilibrium point(s) and whether they are stable, unstable or neutral.





    3. The attempt at a solution

    A. E=(36/5)(q/L^2)

    B. Both are equal a = 2qE/m = qE/(m/2)

    D. 0 at top of square going in towards B from + and towards B for -
     
  2. jcsd
  3. Jan 22, 2014 #2
    What equation are you using for electric field strength?
     
  4. Jan 22, 2014 #3
    E=k(q/R^2)

    and adding
     
  5. Jan 22, 2014 #4
    I got that from the bottom charges R=(sqrt(5)/2)L

    and the other two cancel out because they go in towards B

    E=k(q/(5/4)L^2)
     
  6. Jan 22, 2014 #5

    lightgrav

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    the E-field contributions by the top charges are both rightward:
    away from +Q , and toward -Q (toward being "negative away"),
    but some of this is cancelled by leftward E-field by the bottom charges.
    What fraction of the E-field contribution by the bottom left -Q is matched by the +Q on bottom right?
     
  7. Jan 22, 2014 #6
    The bottom left -Q and +Q would result in a leftward towards upper +Q but I thought the bottom ones would cancel out completely?
     
  8. Jan 22, 2014 #7

    lightgrav

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    away from the positive and toward the negative; only their vertical components cancel ... their horizontals double.
     
  9. Jan 22, 2014 #8
    so their force added would be 8kq/5L^2 .. ?
     
  10. Jan 22, 2014 #9

    lightgrav

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    each E-field contribution is 4/5 kQ/L² , but that is along the diagonal. You need to find the horizontal component.
     
  11. Jan 22, 2014 #10
    horizontal component: 2(4/5)(kQ/L^2)cos(180)+2(4/5)(kQ/L^2)cos(0)
     
  12. Jan 22, 2014 #11

    lightgrav

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    Look at the triangle that you used to find the diagonal distance. The E-field contribution is along that diagonal.
    The negative's contribution is mostly down, but a little bit leftward ... left is L/2 compared to the diagonal L√5 /2 ... that is, the diagonal contribution is √5 times as long as the horizontal part that survives.
     
  13. Jan 22, 2014 #12
    So would the x component be:

    Lcos(180)+Lcos(0)?
    Ok (duh) it's running from b to the top charge(l/2)
    Thanks for your help I never understand the x and y components
     
  14. Jan 22, 2014 #13

    lightgrav

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    /
    /|
    / |
    /_|
    the Ex/E ratio is the same as the Lx/R ratio ... each = sin theta (the skinny angle)
    so Ex = E Lx/R , leftward (negative)
    The E-field contributions from the TOP charges are at 0° (both are rightward)
     
    Last edited: Jan 22, 2014
  15. Jan 22, 2014 #14

    lightgrav

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    |\
    | \ E from bottom right +Q , upward
    | \
    |__\
    Ex

    sin theta = Ex/E , but also sin theta = ½L /R ... so Ex = ½L /R * E
     
  16. Jan 22, 2014 #15
    Okay so add ((L/2)/R)xE and EL/R then multiply by cos(180) + (8kq/5L^2) ?
     
  17. Jan 22, 2014 #16
    Or is the component 1/2L/R *E ?
     
  18. Jan 22, 2014 #17

    lightgrav

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    yes, pointing leftward, if you put the correct expression in for E. and R.
    then you add these (subtract, actually) to the top Q's E-fields ; don't forget those, because they're much bigger!
     
  19. Jan 22, 2014 #18
    Yes to the first post

    Correction at end:: + (8kq/5L^2)cos(0)


    Or two the Ex=(1/2)L/R * E ???
     
  20. Jan 22, 2014 #19

    lightgrav

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    where did the 8 in the 8/5 come from?

    don't confuse yourself by writing "cos(180)" when you mean leftward.
    the triangle base is ½ ; the diagonal is 1.118 , so the base is .5/1.118 times as big.
    the same ratio applies to the red arrows (E-field contributions)
    http://www.science.marshall.edu/foltzc/addEsq.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
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