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elica85

  • 2 years ago

What is the magnitude of the force on the -10 charge in the figure ?

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  1. elica85
    • 2 years ago
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    |dw:1327758982728:dw|

  2. elica85
    • 2 years ago
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    the resultant force should be...|dw:1327759214456:dw| something like that

  3. elica85
    • 2 years ago
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    broke down x and y components for force on + on - and - on - and then added...not getting right answer

  4. fryad_muhammed
    • 2 years ago
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    450*10^-9 N

  5. elica85
    • 2 years ago
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    can you explain?

  6. elica85
    • 2 years ago
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    it's not the right answer

  7. fryad_muhammed
    • 2 years ago
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    why?

  8. Shayaan_Mustafa
    • 2 years ago
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    it is not easy. it is a vector quantity. therefore you must have to consider directions.

  9. Shayaan_Mustafa
    • 2 years ago
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    you have to wait until i solve it for you. ok elicia85?

  10. elica85
    • 2 years ago
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    sure and can you please show steps??

  11. Shayaan_Mustafa
    • 2 years ago
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    yes i will show steps to you. wait. i am here.

  12. elica85
    • 2 years ago
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    ok thank you

  13. Shayaan_Mustafa
    • 2 years ago
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    you are welcome.

  14. Shayaan_Mustafa
    • 2 years ago
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    here is diagram first look at this and understand this.

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  15. Shayaan_Mustafa
    • 2 years ago
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    is this pic making sense to you? if i do anything wrong here then kindly point out me. you too faryad, hei, elicia85 and an unknown person.

  16. elica85
    • 2 years ago
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    looks good

  17. fryad_muhammed
    • 2 years ago
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    thank you Shayan

  18. Shayaan_Mustafa
    • 2 years ago
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    wait i have to go and just come in 2min

  19. Shayaan_Mustafa
    • 2 years ago
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    hello. it is taking too much time for me. lets solve step by step here. ok. force of +15nC will be represented by F15 force of -10nC will be represented by F10 force of -5nC will be represented by F5 ok.

  20. Shayaan_Mustafa
    • 2 years ago
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    now as you can see. on -10nC there are two forces acting on it. 1)F15 which is attractive and at an angle of 45 2)F5 which repulsive.

  21. Shayaan_Mustafa
    • 2 years ago
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    F15 can be resolve into factors. along x-axis and along y-axis.

  22. Shayaan_Mustafa
    • 2 years ago
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    forces along y-axis can be calculated as, F15y=(kq1q2)/r² sin45 and F5y=(kq1q2)/r² forces along x-axis can be calculated as, F15x=(kq1q2)/r² cos45

  23. Shayaan_Mustafa
    • 2 years ago
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    finally. sum forces and you will get your answer.

  24. Shayaan_Mustafa
    • 2 years ago
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    have you got it? it is done.

  25. elica85
    • 2 years ago
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    it's exactly what i did but i will try again

  26. Shayaan_Mustafa
    • 2 years ago
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    can you show your solution?

  27. elica85
    • 2 years ago
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    well you still have to add the F5 on F10 don't you? which will only consist of y direction force

  28. elica85
    • 2 years ago
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    my answer was 4.07*10^-3

  29. Shayaan_Mustafa
    • 2 years ago
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    well you still have to add the F5 on F10 don't you? which will only consist of y direction force on y-axis: F10y=F15sin45 - F5 on x-axis: F10x= -F15x finally, \[F _{10}=\sqrt{(F _{10y})^{2}+(F _{10x})^{2}}\] This is complete.

  30. Shayaan_Mustafa
    • 2 years ago
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    any question?

  31. fryad_muhammed
    • 2 years ago
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    shayan can you help me?

  32. Shayaan_Mustafa
    • 2 years ago
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    of course fryad. although we are brothers. arn't we?

  33. elica85
    • 2 years ago
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    thank you i'll try again!

  34. Shayaan_Mustafa
    • 2 years ago
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    you are ever welcome. Regards, Read my Profile. :D

  35. fryad_muhammed
    • 2 years ago
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    yes ,sure thank you , please prove this for me A.B=cos& |A|.|B|

  36. Shayaan_Mustafa
    • 2 years ago
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    ok w8. let me analyze.

  37. Shayaan_Mustafa
    • 2 years ago
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    The three vectors above form the triangle AOB. using Law of Cosines , \[|a-b|^{2}=|a|^{2}+|b|^{2}-2|a||b|\cos \theta\] using properties of dot product. \[|a-b|^{2}=(a-b).(a-b)\] \[=a.a-a.b-b.a+b.b\] \[=|a|^{2}+|b|^{2}-2a.b\] original becomes, \[|a-b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta\] and, \[|a|^{2}-2a.b+|b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta\] now cancel terms which are same and you will got \[a.b=|a||b|\cos \theta\]

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  38. Shayaan_Mustafa
    • 2 years ago
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    that is all. Regards, Read my Profile. :D

  39. fryad_muhammed
    • 2 years ago
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    zor zor supas aziyatm dait

  40. Shayaan_Mustafa
    • 2 years ago
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    kia?

  41. Shayaan_Mustafa
    • 2 years ago
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    what is this language?

  42. fryad_muhammed
    • 2 years ago
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    what is your language?

  43. Shayaan_Mustafa
    • 2 years ago
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    urdu and english.

  44. fryad_muhammed
    • 2 years ago
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    I am kurdish

  45. Shayaan_Mustafa
    • 2 years ago
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    ok. now good luck. take care everyone.

  46. fryad_muhammed
    • 2 years ago
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    I am beginner in physic,thus I want help me for any time if you can my brother?

  47. Shayaan_Mustafa
    • 2 years ago
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    sure sure. i am often available here. you can post your question.

  48. fryad_muhammed
    • 2 years ago
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    That is all kind of you for every thing.can I recognize you in facebook?

  49. Jemurray3
    • 2 years ago
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    Just skimming through this.... You should recognize that the angle is not 45 degrees because it's a rectangle, not a square, so the values above would not yield the correct answer.

  50. elica85
    • 2 years ago
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    good point, i wasn't using 45 though, just used the measurements given. i still need to resolve, haven't got to it yet

  51. Shayaan_Mustafa
    • 2 years ago
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    elica85 are you here?

  52. Shayaan_Mustafa
    • 2 years ago
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    oh yes. so angle is not 45degree. it can be calculated as, \[\tan^{-1} (0.01/0.03)=18.43degrees.\] Kindly use this angle instead of 45.

  53. Shayaan_Mustafa
    • 2 years ago
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    thanks to Jemurray3.

  54. Jemurray3
    • 2 years ago
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    elica I assume you are adding the components separately and then using the Pythagorean Theorem to calculate the magnitude of the resultant, right?

  55. Shayaan_Mustafa
    • 2 years ago
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    hi jemurray3

  56. Jemurray3
    • 2 years ago
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    Hi there Shayaan

  57. Shayaan_Mustafa
    • 2 years ago
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    http://openstudy.com/study#/updates/4f245a39e4b0a2a9c2667f2f We need you in mathematics section. kindly help us.

  58. elica85
    • 2 years ago
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    thank you everyone, i did get the right answer finally. i was confused about +/-. i thought the - charge on negative would make it negative direction going down y, making it double negative...so i was just getting calculations wrong. it was 18 degrees ^^

  59. Shayaan_Mustafa
    • 2 years ago
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    you are ever welcome. we are made for help. pleasure.

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