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Shhot

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

  • 8 months ago
  • 8 months ago

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  1. Shhot
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    figure attached:

    • 8 months ago
  2. theEric
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    Hi! So, do you have any idea where to start?

    • 8 months ago
  3. Shhot
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    no clue, not sure if q has anything to do with it.

    • 8 months ago
  4. Shhot
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    is the q just to help calculate lengths or does it impact the force?

    • 8 months ago
  5. theEric
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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?

    • 8 months ago
  6. theEric
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    Problems don't usually include unnecessary information.

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

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

    • 8 months ago
  9. theEric
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    (And they might be answered by you, in a moment!)

    • 8 months ago
  10. Shhot
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    the 10nC is pulling down and to the right, and q is pulling down and to the right.

    • 8 months ago
  11. theEric
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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....

    • 8 months ago
  12. Shhot
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    ...must be equal. I understand, but do i specifically need to find q?

    • 8 months ago
  13. theEric
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    It says I'm typing but I'm not... You finish my earlier sentence!

    • 8 months ago
  14. theEric
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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!

    • 8 months ago
  15. theEric
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    10nc*, not -10nC.

    • 8 months ago
  16. theEric
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    C*

    • 8 months ago
  17. theEric
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    nC*

    • 8 months ago
  18. theEric
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    Ah...

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

    • 8 months ago
  20. theEric
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    Solve for q, since it will be the only unknown.

    • 8 months ago
  21. Shhot
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    I dont get what you mean...

    • 8 months ago
  22. xxAshxx
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    What?

    • 8 months ago
  23. Shhot
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    sovling for q, How do I go about doing that? Unsure where to start...

    • 8 months ago
  24. theEric
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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?

    • 8 months ago
  25. theEric
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    And then you can spit it into horizontal and vertical components?

    • 8 months ago
  26. Shhot
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    yep, can do that. Its just finding q and then F

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

    • 8 months ago
  28. theEric
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    No! Wait! I did that wrong! One change.

    • 8 months ago
  29. theEric
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    \[F_{10nC,\ x}+F_{q,\ x}=0\Rightarrow F_{q,\ x}=-F_{10nC,\ x}\]

    • 8 months ago
  30. Shhot
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    still lost... do I use the entire bottom length?

    • 8 months ago
  31. Shhot
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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?

    • 8 months ago
  32. theEric
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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.

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

    • 8 months ago
  34. theEric
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    Yep!

    • 8 months ago
  35. Shhot
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    ok

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

    • 8 months ago
  37. Shhot
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    ok, starting to get clearer.

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

    • 8 months ago
  39. Shhot
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    yes

    • 8 months ago
  40. Shhot
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    For q, I got 3e-8C.

    • 8 months ago
  41. theEric
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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?

    • 8 months ago
  42. theEric
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    If so, you can solve for \(q\) after you find the distances.

    • 8 months ago
  43. Shhot
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    ahh, now i'm lost again...

    • 8 months ago
  44. theEric
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    I used substitution.

    • 8 months ago
  45. Shhot
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    Could I just use trig to find the distances?

    • 8 months ago
  46. theEric
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    Yup! You'll need to. I don't see how to get around it, at least.

    • 8 months ago
  47. theEric
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    Do you see what you need to do?

    • 8 months ago
  48. theEric
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    5cm is the total distance between the bottom charges, right?...

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

    • 8 months ago
  50. theEric
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    The \(r\) you use to calculate a force between two charges is the distance between those charges.

    • 8 months ago
  51. Shhot
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    Along the diagonal? or (since x-components cancel) the y-component

    • 8 months ago
  52. theEric
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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.

    • 8 months ago
  53. Shhot
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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?

    • 8 months ago
  54. Shhot
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    To find q, I have Fq(cos(30) = F10nC(cos(60)

    • 8 months ago
  55. theEric
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    Yep! And yep.

    • 8 months ago
  56. theEric
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    Except they're not quite equal.

    • 8 months ago
  57. theEric
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    Equal in magnitude, opposite in direction. So they cancel.

    • 8 months ago
  58. Shhot
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    yes,F10nC - Fq = 0. I just moved them.

    • 8 months ago
  59. Shhot
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    q=1.7e-8C

    • 8 months ago
  60. theEric
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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.

    • 8 months ago
  61. Shhot
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    I got 1.7e-4 N for total force

    • 8 months ago
  62. theEric
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    I still didn't find the distances.. And then I did more math that became jumbled.

    • 8 months ago
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