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Shhot

  • 2 years ago

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

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  1. Shhot
    • 2 years ago
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    figure attached:

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

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

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

  5. theEric
    • 2 years ago
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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?

  6. theEric
    • 2 years ago
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    Problems don't usually include unnecessary information.

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

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

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

  10. Shhot
    • 2 years ago
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    the 10nC is pulling down and to the right, and q is pulling down and to the right.

  11. theEric
    • 2 years ago
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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....

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

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

  14. theEric
    • 2 years ago
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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!

  15. theEric
    • 2 years ago
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    10nc*, not -10nC.

  16. theEric
    • 2 years ago
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    C*

  17. theEric
    • 2 years ago
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    nC*

  18. theEric
    • 2 years ago
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    Ah...

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

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

  21. Shhot
    • 2 years ago
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    I dont get what you mean...

  22. xxAshxx
    • 2 years ago
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    What?

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

  24. theEric
    • 2 years ago
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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?

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

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

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

  28. theEric
    • 2 years ago
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    No! Wait! I did that wrong! One change.

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

  30. Shhot
    • 2 years ago
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    still lost... do I use the entire bottom length?

  31. Shhot
    • 2 years ago
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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?

  32. theEric
    • 2 years ago
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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.

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

  34. theEric
    • 2 years ago
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    Yep!

  35. Shhot
    • 2 years ago
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    ok

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

  37. Shhot
    • 2 years ago
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    ok, starting to get clearer.

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

  39. Shhot
    • 2 years ago
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    yes

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

  41. theEric
    • 2 years ago
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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?

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

  43. Shhot
    • 2 years ago
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    ahh, now i'm lost again...

  44. theEric
    • 2 years ago
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    I used substitution.

  45. Shhot
    • 2 years ago
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    Could I just use trig to find the distances?

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

  47. theEric
    • 2 years ago
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    Do you see what you need to do?

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

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

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

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

  52. theEric
    • 2 years ago
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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.

  53. Shhot
    • 2 years ago
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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?

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

  55. theEric
    • 2 years ago
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    Yep! And yep.

  56. theEric
    • 2 years ago
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    Except they're not quite equal.

  57. theEric
    • 2 years ago
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    Equal in magnitude, opposite in direction. So they cancel.

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

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

  60. theEric
    • 2 years ago
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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.

  61. Shhot
    • 2 years ago
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    I got 1.7e-4 N for total force

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

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