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osanseviero

  • one year ago

Electricity problem (Coulomb's Law)

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  1. osanseviero
    • one year ago
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    |dw:1439671502885:dw|

  2. osanseviero
    • one year ago
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    If the net force is 4* 10^-6 N, which is the value of q3?

  3. anonymous
    • one year ago
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    Is q3 attracting with q1? and is q3 repelling with q2? I need to know if the point charges are positive or negative.

  4. osanseviero
    • one year ago
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    q1 is negative, q2 is positive, so it should go to the right.

  5. osanseviero
    • one year ago
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    I did F12 + F23 = Fn Then I wrote all the value of each one, and tried to get out q3

  6. anonymous
    • one year ago
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    and what charge is q3?

  7. osanseviero
    • one year ago
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    That is what the problem is asking

  8. anonymous
    • one year ago
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    We need to know charge of q3 to know whether it goes left of q1 or toward q1 and also whether it goes toward q2 or right of q2

  9. anonymous
    • one year ago
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    screenshot the problem.

  10. osanseviero
    • one year ago
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    It is in Spanish

  11. osanseviero
    • one year ago
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    P1. Se ordenan tres cargas puntuales a lo largo del eje de las x. La carga q1=-4.5nC está en x=0.2m, y la carga q2 = 2.5 nC, en x=-0.3m. Hay una carga puntual positiva q3 en el origen. a) ¿Cuál debe ser el valor de q3 para que la fuerza neta sobre esta carga puntual tenga una magnitud de 4x10-6 N? Traduction: There are 3 charges in the x axis. Then they give all the values. What is the value to the force be 4x10^-6 N?

  12. osanseviero
    • one year ago
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    Since the force is positive, I guess that it is going to the right

  13. anonymous
    • one year ago
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    Um. I google translated it. Three point load along the x axis are arranged . Q1 = -4.5nC the load is at x = 0.2m , and q2 = 2.5 nC charge at x = -0.3m . There is a positive point charge at the origin q3 . a) What should be the value of q3 to the net force on the point charge has a magnitude of 4x10-6 N ? Translation : There are 3 charges in the x axis . Then They give all the values ​​. What is the value to the force be 4x10 ^ -6 N ?

  14. anonymous
    • one year ago
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    You see where it says "There is a positive point charge at the origin q3?"

  15. osanseviero
    • one year ago
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    Oh, sorry :/

  16. anonymous
    • one year ago
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    I can still figure it out whether is postie or not by knowing the net force and distance btw. But it takes longer.

  17. anonymous
    • one year ago
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    Let be k be the electric force constant. Since q2 and q3 are positive, they repel, so q3 goes to the right. Since q3 and q1 are opposite charged of each other, q3 attracts q1 so it gets near q1 (goes tot he right also.)

  18. anonymous
    • one year ago
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    \[\frac{ k*q1q3 }{ (0.2m)^2 }+\frac{ kq2q3 }{ (0.3)^2 }=4*10^{-6}N\] where q1=4.5nC or 0.0000000045 C and q2=2.5nC=0.0000000025 C use the decimals it make it faster and plug in k

  19. osanseviero
    • one year ago
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    \[q3 = \frac{ 4 \times 10^{-6} }{ 9\times 10 ^{9} } \times \left( \frac{ r1^{2} }{ q1 } + \frac{ r2^{2} }{ q2 }\right)\] That looks right?

  20. anonymous
    • one year ago
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    I got \[q3(\frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r2^2 })=4x10^{-6}N\]

  21. anonymous
    • one year ago
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    4*10^-6

  22. anonymous
    • one year ago
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    \[q3=\frac{ 4*10^{-6} N}{ \frac{ k*q1 }{ r1^2 }+\frac{ k*q2 }{ r^2 } }\] let r1 be distance between q1 and q3 and r2 be distance between q2 and q3

  23. osanseviero
    • one year ago
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    Wooh got the result. 3.16nC

  24. anonymous
    • one year ago
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    forgot to write r2^2 on the second denominator fraction.

  25. anonymous
    • one year ago
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    Using your equation?

  26. osanseviero
    • one year ago
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    Nope, I did it again

  27. osanseviero
    • one year ago
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    You did it right

  28. anonymous
    • one year ago
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    Let me calculate to make sure.

  29. anonymous
    • one year ago
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    Yes 3.2nC

  30. anonymous
    • one year ago
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    Close enough.

  31. osanseviero
    • one year ago
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    Yep, calculated it again and got 3.2. Thanks a lot for all the help

  32. anonymous
    • one year ago
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    Ye, no problem.

  33. anonymous
    • one year ago
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    Just remember that net force is change in force in the x and y component. It is useful for the one that goes diagonal.

  34. osanseviero
    • one year ago
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    Yep, that is the next problem

  35. osanseviero
    • one year ago
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    If I want the net force to be 0, I would just do F13 = F23 and use r1 = r and r2 = (0.5 - r). I need to get r

  36. anonymous
    • one year ago
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    It depends on the force to set them equal to each other If both are pushing left or right, you cannot set them equal to each other. If there are forces on the left and right and net force is zero, you can make the opposite forces equal to each other. Yea r1=r and r2=(0.5-r) is important to use if you do not know the distance of one of the charges and know the other one.

  37. osanseviero
    • one year ago
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    Ok, thanks :)

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