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woleraymond Group Title

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.

  • 2 years ago
  • 2 years ago

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  1. woleraymond Group Title
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    @Algebraic! any clues

    • 2 years ago
  2. Algebraic! Group Title
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    |dw:1348695082199:dw|

    • 2 years ago
  3. Algebraic! Group Title
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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

    • 2 years ago
  4. Algebraic! Group Title
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    is it clear?

    • 2 years ago
  5. woleraymond Group Title
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    not clear at all

    • 2 years ago
  6. Algebraic! Group Title
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    the forces are equal in magnitude and opposite in direction.

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

    • 2 years ago
  8. Algebraic! Group Title
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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

    • 2 years ago
  9. Algebraic! Group Title
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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

    • 2 years ago
  10. Algebraic! Group Title
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    solve the quadratic for x

    • 2 years ago
  11. Algebraic! Group Title
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    clear now?

    • 2 years ago
  12. woleraymond Group Title
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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?

    • 2 years ago
  13. Algebraic! Group Title
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    +4e charge is at x=0; +2e charge is at x=.4

    • 2 years ago
  14. Algebraic! Group Title
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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

    • 2 years ago
  15. Algebraic! Group Title
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    |dw:1348774582953:dw|

    • 2 years ago
  16. woleraymond Group Title
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    there is no 4 its 40..also the (0,0) isnt the x,y pairs?

    • 2 years ago
  17. Algebraic! Group Title
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    .4

    • 2 years ago
  18. Algebraic! Group Title
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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

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

    • 2 years ago
  20. Algebraic! Group Title
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    ok?

    • 2 years ago
  21. woleraymond Group Title
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    so u mean the 3rd charge is in the middle of both charge?

    • 2 years ago
  22. Algebraic! Group Title
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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...

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

    • 2 years ago
  24. Algebraic! Group Title
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    4e/x^2 = 2e/(.4-x)^2

    • 2 years ago
  25. Algebraic! Group Title
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    cross multiply

    • 2 years ago
  26. Algebraic! Group Title
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    4e(.4-x)^2 = 2ex^2

    • 2 years ago
  27. woleraymond Group Title
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    x is 2.3

    • 2 years ago
  28. woleraymond Group Title
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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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