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Shhot

  • one year 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
    • one year ago
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    figure attached:

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

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

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

  5. theEric
    • one year 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
    • one year ago
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    Problems don't usually include unnecessary information.

  7. Shhot
    • one year 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
    • one year ago
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    So, conceptually, why does the force go down? Your other questions will be answered momentarily.

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

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

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

  14. theEric
    • one year 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
    • one year ago
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    10nc*, not -10nC.

  16. theEric
    • one year ago
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    C*

  17. theEric
    • one year ago
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    nC*

  18. theEric
    • one year ago
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    Ah...

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

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

  22. xxAshxx
    • one year ago
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    What?

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

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

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

  27. theEric
    • one year 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
    • one year ago
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    No! Wait! I did that wrong! One change.

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

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

  31. Shhot
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    Yep!

  35. Shhot
    • one year ago
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    ok

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

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

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

  39. Shhot
    • one year ago
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    yes

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

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

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

  44. theEric
    • one year ago
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    I used substitution.

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

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

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

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

  49. Shhot
    • one year 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
    • one year ago
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    The \(r\) you use to calculate a force between two charges is the distance between those charges.

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

  52. theEric
    • one year 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
    • one year 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
    • one year ago
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    To find q, I have Fq(cos(30) = F10nC(cos(60)

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

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

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

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

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

  60. theEric
    • one year 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
    • one year ago
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    I got 1.7e-4 N for total force

  62. theEric
    • one year 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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