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Shhot Group Title

The force on the -1.0nC charge is as shown in the figure. What is the magnitude of this force?

  • one year ago
  • one year ago

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  1. Shhot Group Title
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    figure attached:

    • one year ago
  2. theEric Group Title
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    Hi! So, do you have any idea where to start?

    • one year ago
  3. Shhot Group Title
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    no clue, not sure if q has anything to do with it.

    • one year ago
  4. Shhot Group Title
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    is the q just to help calculate lengths or does it impact the force?

    • one year ago
  5. theEric Group Title
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    Well, it does if it has an effect on the \(-1\ [nC]\) charge! It will definitely have an effect on the force. Here's a question that will get you started: why is the force straight down? And then, what does that mean?

    • one year ago
  6. theEric Group Title
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    Problems don't usually include unnecessary information.

    • one year ago
  7. Shhot Group Title
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    I know why the charge goes down conceptually, but do I have to calculate q and if so, how?

    • one year ago
  8. theEric Group Title
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    So, conceptually, why does the force go down? Your other questions will be answered momentarily.

    • one year ago
  9. theEric Group Title
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    (And they might be answered by you, in a moment!)

    • one year ago
  10. Shhot Group Title
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    the 10nC is pulling down and to the right, and q is pulling down and to the right.

    • one year ago
  11. theEric Group Title
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    "the 10nC is pulling down and to the" \(left\), "and q is pulling down and to the right." I'm sure you thought that, but mistyped. And for there to be no horizontal component of force on the -1nC charge, the horizontal forces from 10nC and q....

    • one year ago
  12. Shhot Group Title
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    ...must be equal. I understand, but do i specifically need to find q?

    • one year ago
  13. theEric Group Title
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    It says I'm typing but I'm not... You finish my earlier sentence!

    • one year ago
  14. theEric Group Title
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    Yep, you're right! And to find the net force on the -1nC charge, you do need \(q\)! It obviously has an impact on the -1nC charge, since it cancels the horizontal force of the -10nC charge! And knowing \(q\) will be necessary to calculate the force on the -1nC charge!

    • one year ago
  15. theEric Group Title
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    10nc*, not -10nC.

    • one year ago
  16. theEric Group Title
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    C*

    • one year ago
  17. theEric Group Title
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    nC*

    • one year ago
  18. theEric Group Title
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    Ah...

    • one year ago
  19. theEric Group Title
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    So the horizontal component of forces from the 10nC and q charges add to zero. Time for some math?

    • one year ago
  20. theEric Group Title
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    Solve for q, since it will be the only unknown.

    • one year ago
  21. Shhot Group Title
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    I dont get what you mean...

    • one year ago
  22. xxAshxx Group Title
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    What?

    • one year ago
  23. Shhot Group Title
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    sovling for q, How do I go about doing that? Unsure where to start...

    • one year ago
  24. theEric Group Title
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    Okay, well, using the formula\[F=k\frac{q_1\ q_2}{r^2}\]you can "find" the force of the 10nC charge on the -1nC charge. Right?

    • one year ago
  25. theEric Group Title
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    And then you can spit it into horizontal and vertical components?

    • one year ago
  26. Shhot Group Title
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    yep, can do that. Its just finding q and then F

    • one year ago
  27. theEric Group Title
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    \[F_{10nC}+F_{q}=0\Rightarrow F_q=F_{10nC}\] \[F_{10nC}=\frac{\left( 10\ [nC]\right)\ \left( -1\ [nC]\right)}{r_{10nC\rightarrow -1nC}}\] \[F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}\]

    • one year ago
  28. theEric Group Title
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    No! Wait! I did that wrong! One change.

    • one year ago
  29. theEric Group Title
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    \[F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=-F_{10nC,\ x}\]

    • one year ago
  30. Shhot Group Title
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    still lost... do I use the entire bottom length?

    • one year ago
  31. Shhot Group Title
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    when you say\[F _{10nC}=\frac{ (10nC)(1nC) }{r10nC \rightarrow-1nC }\] Is that the length along the diagonal or the x-component?

    • one year ago
  32. theEric Group Title
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    Diagonal, sorry. It must be the distance between the 10nC and -1nC charges if we want to calculate the force between them. Have you tried drawing a force diagram for forces on the -1nC charge? They're very helpful! I'll post mine momentarily.

    • one year ago
  33. Shhot Group Title
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    yes, so i use the diagonals for both the 10nC, 1nC and q,1nC when finding value of q?

    • one year ago
  34. theEric Group Title
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    Yep!

    • one year ago
  35. Shhot Group Title
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    ok

    • one year ago
  36. theEric Group Title
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    The horizontal forces cancel each other out. \( F_q cos(30^\circ) = -F_{10nC}cos(60^\circ)\)

    • one year ago
  37. Shhot Group Title
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    ok, starting to get clearer.

    • one year ago
  38. theEric Group Title
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    \[F_q = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\] Still good?

    • one year ago
  39. Shhot Group Title
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    yes

    • one year ago
  40. Shhot Group Title
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    For q, I got 3e-8C.

    • one year ago
  41. theEric Group Title
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    \[F_{q}=\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}}\] So we substitute \(F_q\)'s value in where \(F_q\) is in\[F_q = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\]to get\[\frac{\left(q\right)\ \left( -1\ [nC]\right)}{r_{q\rightarrow -1nC}} = -F_{10nC}\frac{cos(60^\circ)}{cos(30^\circ) }\] Still with me?

    • one year ago
  42. theEric Group Title
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    If so, you can solve for \(q\) after you find the distances.

    • one year ago
  43. Shhot Group Title
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    ahh, now i'm lost again...

    • one year ago
  44. theEric Group Title
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    I used substitution.

    • one year ago
  45. Shhot Group Title
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    Could I just use trig to find the distances?

    • one year ago
  46. theEric Group Title
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    Yup! You'll need to. I don't see how to get around it, at least.

    • one year ago
  47. theEric Group Title
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    Do you see what you need to do?

    • one year ago
  48. theEric Group Title
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    5cm is the total distance between the bottom charges, right?...

    • one year ago
  49. Shhot Group Title
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    I know how to find the distances. Its just figuring out which distance goes into r for\[F=\frac{ Kq1q2 }{ r^2}\]

    • one year ago
  50. theEric Group Title
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    The \(r\) you use to calculate a force between two charges is the distance between those charges.

    • one year ago
  51. Shhot Group Title
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    Along the diagonal? or (since x-components cancel) the y-component

    • one year ago
  52. theEric Group Title
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    Along the diagonal, for sure. To get any one component alone, you have to get the total force and break it up. If you used just the \(x\) distance in your formula, you'd be calculating the force as if the charges themselves were that far away.

    • one year ago
  53. Shhot Group Title
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    oh, so use the diagonal distance for r and then multiply \[\frac{ Kq1q2 }{ r^2 }\] by either sin/cos(theta) to get x or y component?

    • one year ago
  54. Shhot Group Title
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    To find q, I have Fq(cos(30) = F10nC(cos(60)

    • one year ago
  55. theEric Group Title
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    Yep! And yep.

    • one year ago
  56. theEric Group Title
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    Except they're not quite equal.

    • one year ago
  57. theEric Group Title
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    Equal in magnitude, opposite in direction. So they cancel.

    • one year ago
  58. Shhot Group Title
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    yes,F10nC - Fq = 0. I just moved them.

    • one year ago
  59. Shhot Group Title
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    q=1.7e-8C

    • one year ago
  60. theEric Group Title
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    F10nC + Fq = 0. The net horizontal force, which is the sum of all horizontal forces, must equal zero. That number seems reasonable! I didn't find the distances.

    • one year ago
  61. Shhot Group Title
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    I got 1.7e-4 N for total force

    • one year ago
  62. theEric Group Title
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    I still didn't find the distances.. And then I did more math that became jumbled.

    • one year ago
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