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woleraymond Group Title

derive an expression for the work done in putting the four charges together as shown below

  • one year ago
  • one year ago

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  1. Algebraic! Group Title
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    |dw:1348178251421:dw|

    • one year ago
  2. woleraymond Group Title
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    |dw:1348178242336:dw|

    • one year ago
  3. Algebraic! Group Title
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    #1 is free...

    • one year ago
  4. Algebraic! Group Title
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    #2 is -qV where V is kq/r so the work is -kq^2/a

    • one year ago
  5. Algebraic! Group Title
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    #3 is -kq^2/a + kq^2/(a*sqrt(2))

    • one year ago
  6. Algebraic! Group Title
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    see where this is going?

    • one year ago
  7. woleraymond Group Title
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    im trying to figgy it

    • one year ago
  8. Algebraic! Group Title
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    |dw:1348178828104:dw|

    • one year ago
  9. Algebraic! Group Title
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    another way to look at it is : \[W =\int\limits_{\infty}^{a} Fdr =\int\limits_{\infty}^{a} k \frac{ qq }{ r ^{2} } dr\]

    • one year ago
  10. Algebraic! Group Title
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    integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course

    • one year ago
  11. Algebraic! Group Title
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    I left signs out because it's easy to see, two opposite charges coming together is negative work and two like charges coming together takes work...

    • one year ago
  12. woleraymond Group Title
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    @Algebraic! note the distance for charge 2 to come the center is \[\sqrt{2a}\]

    • one year ago
  13. Algebraic! Group Title
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    a

    • one year ago
  14. Algebraic! Group Title
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    nothing is coming to the 'center'

    • one year ago
  15. RaphaelFilgueiras Group Title
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    the work is the total energy to assembly this configuration so: \[w=1/4.\pi.\epsilon0[q1q2/a+q1q3/\sqrt{2a}+q1.q4/a+q2q3/a+q2q4/\sqrt{2a}+q3q4/a]\]

    • one year ago
  16. Algebraic! Group Title
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    you're asked to assemble the charges into the config. shown... that means moving #2 from infinity to a against the potential of #1

    • one year ago
  17. RaphaelFilgueiras Group Title
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    |dw:1348180508889:dw|

    • one year ago
  18. RaphaelFilgueiras Group Title
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    @woleraymond get it?

    • one year ago
  19. Algebraic! Group Title
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    @RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?

    • one year ago
  20. Algebraic! Group Title
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    I see.

    • one year ago
  21. woleraymond Group Title
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    @RaphaelFilgueiras and @Algebraic! u guys are good

    • one year ago
  22. Algebraic! Group Title
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    dead wrong

    • one year ago
  23. Algebraic! Group Title
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    which direction is F in k(q)(-q) /r^2 and which direction is F in k(q)(q) /r^2 ??

    • one year ago
  24. woleraymond Group Title
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    @RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other

    • one year ago
  25. RaphaelFilgueiras Group Title
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    the work is the energy of configuration

    • one year ago
  26. Algebraic! Group Title
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    lol do the work intergal @RaphaelFilgueiras

    • one year ago
  27. RaphaelFilgueiras Group Title
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    if you get a negative energy is a spontaneous config

    • one year ago
  28. Algebraic! Group Title
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    |dw:1348180925907:dw|

    • one year ago
  29. Algebraic! Group Title
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    yeah, two opposite charges move together spontaneously...

    • one year ago
  30. RaphaelFilgueiras Group Title
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    no you have to put it together so you will do work

    • one year ago
  31. Algebraic! Group Title
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    in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow

    • one year ago
  32. RaphaelFilgueiras Group Title
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    two positives charges???will atract each other?

    • one year ago
  33. woleraymond Group Title
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    yep @RaphaelFilgueiras work done from + to - should be -ve and + to + be +ve...what ypu think

    • one year ago
  34. Algebraic! Group Title
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    mother of god http://i.imgur.com/jPZTm.gif

    • one year ago
  35. RaphaelFilgueiras Group Title
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    my halliday must be wrong

    • one year ago
  36. Algebraic! Group Title
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    you're doing it wrong

    • one year ago
  37. RaphaelFilgueiras Group Title
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    ok,for sure

    • one year ago
  38. Algebraic! Group Title
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    dude, just do the work integral, check the sign, end of story.

    • one year ago
  39. RaphaelFilgueiras Group Title
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    my way is rigth too

    • one year ago
  40. Algebraic! Group Title
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    different answers, however. How shall we reconcile this???

    • one year ago
  41. Algebraic! Group Title
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    you're doing it for all LIKE charges... these aren't all like charges....

    • one year ago
  42. RaphaelFilgueiras Group Title
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    no you must put the signal in formula!!!!

    • one year ago
  43. RaphaelFilgueiras Group Title
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    q1=-q

    • one year ago
  44. RaphaelFilgueiras Group Title
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    lol

    • one year ago
  45. Algebraic! Group Title
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    @Mikael

    • one year ago
  46. Algebraic! Group Title
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    now you're in trouble...

    • one year ago
  47. Algebraic! Group Title
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    where you from @RaphaelFilgueiras ?

    • one year ago
  48. Algebraic! Group Title
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    brazil?

    • one year ago
  49. RaphaelFilgueiras Group Title
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    yes

    • one year ago
  50. Algebraic! Group Title
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    is your Halliday in Portuguese?

    • one year ago
  51. RaphaelFilgueiras Group Title
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    • one year ago
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  52. Algebraic! Group Title
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    yep

    • one year ago
  53. Algebraic! Group Title
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    now tell me +q * -q = +q *+q

    • one year ago
  54. Algebraic! Group Title
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    better yet, find the page in Halliday that says that...

    • one year ago
  55. Algebraic! Group Title
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    inb4 I couldn't find it but I know I saw it once

    • one year ago
  56. RaphaelFilgueiras Group Title
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    q1=-q;q2=+q;q3=-q;q4=+q

    • one year ago
  57. Algebraic! Group Title
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    you got that much right... continue...

    • one year ago
  58. Algebraic! Group Title
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    pretty much boils down to whether you think these are equal: \[\int\limits_{\infty}^{R} \frac{ k(q)(-q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]

    • one year ago
  59. Algebraic! Group Title
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    or opposite.

    • one year ago
  60. RaphaelFilgueiras Group Title
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    • one year ago
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  61. Algebraic! Group Title
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    you're conceding I take it?

    • one year ago
  62. RaphaelFilgueiras Group Title
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    nope,this was my first answer

    • one year ago
  63. RaphaelFilgueiras Group Title
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    what is yours?

    • one year ago
  64. Algebraic! Group Title
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    omg you think that that page you linked is evidence IN your favor??

    • one year ago
  65. Algebraic! Group Title
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    your answer is all positive terms...

    • one year ago
  66. RaphaelFilgueiras Group Title
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    no i put q1.q2

    • one year ago
  67. RaphaelFilgueiras Group Title
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    and said q1=-q ;q2=q....

    • one year ago
  68. Algebraic! Group Title
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    there are no q1's or q2's on the sketch

    • one year ago
  69. Algebraic! Group Title
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    there's q and -q

    • one year ago
  70. Algebraic! Group Title
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    and you've already said you think the potential is positive independent of charge...

    • one year ago
  71. RaphaelFilgueiras Group Title
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    ok bye

    • one year ago
  72. Algebraic! Group Title
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    bye.

    • one year ago
  73. woleraymond Group Title
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    @Algebraic! @RaphaelFilgueiras please what is the way forward on this

    • one year ago
  74. Algebraic! Group Title
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    I did 3/4 terms of the solution, do you understand how we get them?

    • one year ago
  75. Algebraic! Group Title
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    all that's left is to do the last charge...

    • one year ago
  76. Algebraic! Group Title
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    hello?

    • one year ago
  77. Algebraic! Group Title
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    RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...

    • one year ago
  78. Algebraic! Group Title
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    are you there @woleraymond ?

    • one year ago
  79. woleraymond Group Title
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    yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did

    • one year ago
  80. Algebraic! Group Title
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    I don't know how else to explain it... the work done is the energy of the configuration... it takes no work to place the first charge... you with me so far?

    • one year ago
  81. woleraymond Group Title
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    yes i agree, it takes no work to place the first charge as it must be -ve same also as the 3rd charge

    • one year ago
  82. Algebraic! Group Title
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    the second charge has to be moved from infinity to a distance of a from the first charge... so \[W =\int\limits_{\infty}^{a} q\frac{ k(-q) }{ r ^{2} }dr\] which is just (-kqq)/a

    • one year ago
  83. Algebraic! Group Title
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    make sense?

    • one year ago
  84. woleraymond Group Title
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    i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also

    • one year ago
  85. Algebraic! Group Title
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    the third charge has to be moved from infinity to a spot a meters from the second charge and a*sqrt(2) from the first charge:

    • one year ago
  86. Algebraic! Group Title
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    brb equation editor is broken

    • one year ago
  87. Algebraic! Group Title
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    this site is buggy

    • one year ago
  88. Algebraic! Group Title
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    \[W = \int\limits_{\infty}^{a} q \frac{ k(-q) }{ r ^{2} } dr +\int\limits_{\infty}^{a \sqrt{2}} -q \frac{ k(-q) }{ r ^{2}}dr\]

    • one year ago
  89. Algebraic! Group Title
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    and so on

    • one year ago
  90. Algebraic! Group Title
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    W for the last charge is moving it to 'a' (meters) from the first charge, 'a*sqrt(2)' from the second charge and 'a' from the 3rd charge

    • one year ago
  91. Algebraic! Group Title
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    got it?

    • one year ago
  92. Algebraic! Group Title
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    "i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also" naw, you put it together bit by bit... first you put in the charge I labeled '1', that's free since there are no charges in place yet, then you bring in the charge I labelled '2', there's only one charge in place so the work is the work done moving it into place against the field of the charge labelled '1', and so on...

    • one year ago
  93. Algebraic! Group Title
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    charge labelled '3' has to be put in place by doing work against the fields of charge 1 and charge 2 .... and charge 4 has to be put in place by doing work against the fields of charges 1, 2 and 3... make sense?

    • one year ago
  94. Algebraic! Group Title
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    @woleraymond hello?

    • one year ago
  95. woleraymond Group Title
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    u making sense...@Algebraic! i got the import..will put every thing together in a bit

    • one year ago
  96. woleraymond Group Title
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    @Algebraic! what expression can we deduce out of this 4 items below #1 is free #2 is -qV where V is kq/r so the work is -kq^2/a #3 is -kq^2/a + kq^2/(a*sqrt(2)) #4 is -kq^2/a + kq^2/(a*sqrt(2)) - kq^2/a

    • one year ago
  97. Algebraic! Group Title
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    yep

    • one year ago
  98. Algebraic! Group Title
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    that's the total work required to assemble them.

    • one year ago
  99. Algebraic! Group Title
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    ok?

    • one year ago
  100. woleraymond Group Title
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    we have to add them up algebraically i guess?

    • one year ago
  101. Algebraic! Group Title
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    yeah

    • one year ago
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