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woleraymond

Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm on the x axis. A third charge e is placed on the straight line joining the two charges so that e is in equilibrium. Find the position of e.

  • one year ago
  • one year ago

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  1. woleraymond
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    @Algebraic! any clues

    • one year ago
  2. Algebraic!
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    |dw:1348695082199:dw|

    • one year ago
  3. Algebraic!
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    q1 (+4e) pushes a positive charge to the right. q2 (+2e) push a positive charge to the left. at some point between them those opposing pushes will be equal in magnitude. sub.s in x and .4-x for r1 and r2 and solve for x

    • one year ago
  4. Algebraic!
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    is it clear?

    • one year ago
  5. woleraymond
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    not clear at all

    • one year ago
  6. Algebraic!
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    the forces are equal in magnitude and opposite in direction.

    • one year ago
  7. Algebraic!
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    | e*E1 |=| e*E2 | E1 = kq1/(r1)^2 E2 = kq2/(r2^2)

    • one year ago
  8. Algebraic!
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    r1 is the distance from the charge at the origin to the location of the point where they 'test' charge is in equilibrium. That distance is x. r2 is the distance from the charge at the .4m to the location of the point where they 'test' charge is in equilibrium. That distance is .4m - x

    • one year ago
  9. Algebraic!
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    E1 = kq1/(r1)^2 = E2 = kq2/(r2)^2 kq1/(r1)^2 = kq2/(r2)^2 kq1/(x)^2 = kq2/(.4-x)^2

    • one year ago
  10. Algebraic!
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    solve the quadratic for x

    • one year ago
  11. Algebraic!
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    clear now?

    • one year ago
  12. woleraymond
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    @Algebraic! not clear..how do you interepret Two charges +4e and +2e are fixed at points (0,0)cm and (0,40)cm..also how u derive the value for r1 and r2..what about the 3rd charge?

    • one year ago
  13. Algebraic!
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    +4e charge is at x=0; +2e charge is at x=.4

    • one year ago
  14. Algebraic!
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    r for the +4e charge is the distance from x=0 to the equilibrium point we're trying to find... call that distance x, r for the +2e charge is then the distance from x=.4 to the equilibrium point we're trying to find so .4-x

    • one year ago
  15. Algebraic!
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    |dw:1348774582953:dw|

    • one year ago
  16. woleraymond
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    there is no 4 its 40..also the (0,0) isnt the x,y pairs?

    • one year ago
  17. Algebraic!
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    .4

    • one year ago
  18. Algebraic!
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    both charges are on the x axis and the equilibrium point is on a line between them, so it's on the x axis too. I don't know why they wrote it like this: "(0,0)cm and (0,40)cm on the x axis." it contradicts itself. so maybe check it to see if that's what the problem actually says. I assumed what they actually meant was both charges on the x axis one at x=0 one at x=.4

    • one year ago
  19. Algebraic!
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    anyway, it's the same problem if those are actually meant to be ordered pairs, because: |dw:1348775056115:dw|

    • one year ago
  20. Algebraic!
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    ok?

    • one year ago
  21. woleraymond
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    so u mean the 3rd charge is in the middle of both charge?

    • one year ago
  22. Algebraic!
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    you have to locate where the third charge is... ...that's the point of the problem. but yes, it's probably on the x-axis, depending on what your problem statement actually says...

    • one year ago
  23. woleraymond
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    how do we solve this kind of quadratic 4e/x^2 = 2e/(4-x)^2

    • one year ago
  24. Algebraic!
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    4e/x^2 = 2e/(.4-x)^2

    • one year ago
  25. Algebraic!
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    cross multiply

    • one year ago
  26. Algebraic!
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    4e(.4-x)^2 = 2ex^2

    • one year ago
  27. woleraymond
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    x is 2.3

    • one year ago
  28. woleraymond
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    @Algebraic! do you believe i just saw the .4 and not 4...thanks a lot

    • one year ago
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