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JenniferSmart1

  • one year ago

I'm adding my vectors incorrectly. Please give me a min to draw it.

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  1. JenniferSmart1
    • one year ago
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    |dw:1360530853167:dw|

  2. JenniferSmart1
    • one year ago
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    \[\vec{F}_{net}=(\vec F_{41}-\vec F_{31}\cos(45))\hat i +(\vec F_{21}-\vec F_{31}\sin(45)) \hat j\]

  3. JenniferSmart1
    • one year ago
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    |dw:1360531227825:dw|

  4. JenniferSmart1
    • one year ago
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    \[F_{net}=\left( \frac{kq_4q_1}{r_{41}^2}-\frac{kq_3q_1}{r_{31}^2}\sin(45)\right) \hat i \left(\frac{kq_2q_1}{r_{21}^2}-\frac{kq_3q_1}{r_{31}^2}\cos(45)\right) \hat j\]

  5. JenniferSmart1
    • one year ago
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    \[q_1=3\times10^{-9}\] \[q_2=9\times10^{-9}\] \[q_3=-9\times10^{-9}\] \[q_4=9\times10^{-9}\]

  6. JenniferSmart1
    • one year ago
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    |dw:1360531777377:dw|

  7. JenniferSmart1
    • one year ago
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    @phi

  8. JenniferSmart1
    • one year ago
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    I get \[(72974.5 N)\hat i\]

  9. JenniferSmart1
    • one year ago
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    which is wrong...

  10. JenniferSmart1
    • one year ago
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    .000392 should be the answer

  11. JenniferSmart1
    • one year ago
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    |dw:1360532089439:dw|

  12. phi
    • one year ago
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    8 orders of magnitude is quite an error

  13. JenniferSmart1
    • one year ago
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    it sure is ...haha

  14. JenniferSmart1
    • one year ago
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    the charges were given in nano coulombs. I converted them to coulombs. \[q_1=+3nC=3\times 10^{-9}\]

  15. JenniferSmart1
    • one year ago
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    k=8.99E9

  16. phi
    • one year ago
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    First, what units is the answer given in ?

  17. JenniferSmart1
    • one year ago
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    Newtons \hat{i}

  18. JenniferSmart1
    • one year ago
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    maybe I read the question wrong

  19. phi
    • one year ago
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    your approach is ok. I would write down the components of each vector

  20. phi
    • one year ago
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    the k 9e9 times 9e-9 gives 81. that still leaves a 3e-9 are you double counting the k ?

  21. JenniferSmart1
    • one year ago
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    |dw:1360532909843:dw| oh my gosh ...yes I think soooo

  22. JenniferSmart1
    • one year ago
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    I'll do it on wolfram... one sec

  23. JenniferSmart1
    • one year ago
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    I found another error...this looks better http://www.wolframalpha.com/input/?i=%28coulombs+constant+*9nC*3nC%29%2F%282cm^2%29++-+++%28coulombs+constant+*-9nC*3nC%29%2F%282.8cm^2%29

  24. JenniferSmart1
    • one year ago
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    silly question... \[F_{31}=F_{31x}+F_{31y}\] is this true?

  25. JenniferSmart1
    • one year ago
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    or is it \[F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}\]

  26. phi
    • one year ago
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    ignoring the units (we can figure those out later) the force from the top left corner is <9*3/4 i 0 j> the force from the bottom right is < 0 i 9*3/4 j> the force from the bottom left is < -9*3/8 * 1/sqrt(2) i - 9*3/8 *1/sqrt(2) j> add up the components to get 27/4( 1- 1/2sqrt(2)) for both i and j components that number is 4.3635 now to get the units 1e-18 C^2 * 9e-9 N-M^2 /C^2 * 1e4 M^2 (to fix the cm) that factor is 9e-5 and our result is 4.3635*9e-5 = 0.0003927 N for both the i and j components

  27. phi
    • one year ago
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    as for your "silly" question. if by F31x you mean < F31x 0> and by F31y < 0 F31y> then yes, F31= F31x+F31y the magnitude of F31 is as you wrote it.

  28. JenniferSmart1
    • one year ago
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    why is "the force from the bottom left is < -9*3/8 * 1/sqrt(2) i" Why did you divide by 8

  29. phi
    • one year ago
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    the distance (diagonal) is 2 sqrt(2). squared you get 4*2= 8 the 1/sqrt(2) is cos(45) (often written as sqrt(2)/2 but we don't need to write it that way)

  30. phi
    • one year ago
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    your 2.8 was just an approximation

  31. phi
    • one year ago
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    I think the resultant force vector is < 0.0003927 i 0.0003927 j> with a magnitude of 0.000555 N pointing in the opposite direction of the diagonal

  32. JenniferSmart1
    • one year ago
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    Yep that's correct...I'm still trying to make sense of it...one second...

  33. JenniferSmart1
    • one year ago
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    \[(9\cdot 3 \cdot\frac 14+(-9)3\cdot \frac 18 \cdot \frac 1{\sqrt2})\hat i\] and for \hat j as well Yep that makes sense

  34. JenniferSmart1
    • one year ago
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    what happened to k

  35. phi
    • one year ago
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    I was ignoring the constants and units, just to get a rough idea of what is going on. I see that both the i and j components are positive and equal to each other. but obviously you have to multiply every term by k, and by 1e-9 (twice! to change nC to C) and by 1e4 cm^2 per meter^2 to get the units correct.

  36. JenniferSmart1
    • one year ago
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    Oh I see

  37. phi
    • one year ago
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    if we use 8.99e9 for k, we match the book's answer 0.000392

  38. JenniferSmart1
    • one year ago
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    shoot me....I continue to get .0000082137 \[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}\]\[-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{0.08m^2}\]

  39. JenniferSmart1
    • one year ago
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    \[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}\]-\[-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{\sqrt0.08m^2}\]

  40. phi
    • one year ago
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    First, I would not write down that expression. I would write down the 3 individual vectors (with components i and j even if 0) Second, your distance squared should be written \( (2e{-2}\ m)^2\)= \( 4e{-4}\ m^2\)

  41. phi
    • one year ago
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    and \( 2 \sqrt{2} \) cm squared is written as 8e-4 m^2

  42. JenniferSmart1
    • one year ago
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    finally!!!1

  43. phi
    • one year ago
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    Good.

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