Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

woleraymond

  • 2 years ago

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

  • This Question is Closed
  1. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1348178251421:dw|

  2. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1348178242336:dw|

  3. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    #1 is free...

  4. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    #2 is -qV where V is kq/r so the work is -kq^2/a

  5. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    #3 is -kq^2/a + kq^2/(a*sqrt(2))

  6. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    see where this is going?

  7. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im trying to figgy it

  8. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1348178828104:dw|

  9. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    another way to look at it is : \[W =\int\limits_{\infty}^{a} Fdr =\int\limits_{\infty}^{a} k \frac{ qq }{ r ^{2} } dr\]

  10. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course

  11. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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...

  12. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Algebraic! note the distance for charge 2 to come the center is \[\sqrt{2a}\]

  13. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    a

  14. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    nothing is coming to the 'center'

  15. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're asked to assemble the charges into the config. shown... that means moving #2 from infinity to a against the potential of #1

  17. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1348180508889:dw|

  18. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @woleraymond get it?

  19. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?

  20. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I see.

  21. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @RaphaelFilgueiras and @Algebraic! u guys are good

  22. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    dead wrong

  23. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    which direction is F in k(q)(-q) /r^2 and which direction is F in k(q)(q) /r^2 ??

  24. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other

  25. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the work is the energy of configuration

  26. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lol do the work intergal @RaphaelFilgueiras

  27. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if you get a negative energy is a spontaneous config

  28. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1348180925907:dw|

  29. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah, two opposite charges move together spontaneously...

  30. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no you have to put it together so you will do work

  31. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow

  32. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    two positives charges???will atract each other?

  33. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep @RaphaelFilgueiras work done from + to - should be -ve and + to + be +ve...what ypu think

  34. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    mother of god http://i.imgur.com/jPZTm.gif

  35. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my halliday must be wrong

  36. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're doing it wrong

  37. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok,for sure

  38. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    dude, just do the work integral, check the sign, end of story.

  39. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my way is rigth too

  40. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    different answers, however. How shall we reconcile this???

  41. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're doing it for all LIKE charges... these aren't all like charges....

  42. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no you must put the signal in formula!!!!

  43. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    q1=-q

  44. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol

  45. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @Mikael

  46. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now you're in trouble...

  47. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    where you from @RaphaelFilgueiras ?

  48. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    brazil?

  49. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  50. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    is your Halliday in Portuguese?

  51. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  52. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yep

  53. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now tell me +q * -q = +q *+q

  54. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    better yet, find the page in Halliday that says that...

  55. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    inb4 I couldn't find it but I know I saw it once

  56. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    q1=-q;q2=+q;q3=-q;q4=+q

  57. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you got that much right... continue...

  58. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    pretty much boils down to whether you think these are equal: \[\int\limits_{\infty}^{R} \frac{ k(q)(-q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]

  59. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or opposite.

  60. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  61. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're conceding I take it?

  62. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope,this was my first answer

  63. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what is yours?

  64. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    omg you think that that page you linked is evidence IN your favor??

  65. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    your answer is all positive terms...

  66. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no i put q1.q2

  67. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and said q1=-q ;q2=q....

  68. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    there are no q1's or q2's on the sketch

  69. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    there's q and -q

  70. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and you've already said you think the potential is positive independent of charge...

  71. RaphaelFilgueiras
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok bye

  72. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    bye.

  73. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Algebraic! @RaphaelFilgueiras please what is the way forward on this

  74. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I did 3/4 terms of the solution, do you understand how we get them?

  75. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    all that's left is to do the last charge...

  76. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hello?

  77. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...

  78. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    are you there @woleraymond ?

  79. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did

  80. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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?

  81. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i agree, it takes no work to place the first charge as it must be -ve same also as the 3rd charge

  82. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  83. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    make sense?

  84. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also

  85. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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:

  86. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    brb equation editor is broken

  87. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this site is buggy

  88. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[W = \int\limits_{\infty}^{a} q \frac{ k(-q) }{ r ^{2} } dr +\int\limits_{\infty}^{a \sqrt{2}} -q \frac{ k(-q) }{ r ^{2}}dr\]

  89. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and so on

  90. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  91. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    got it?

  92. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    "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...

  93. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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?

  94. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @woleraymond hello?

  95. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    u making sense...@Algebraic! i got the import..will put every thing together in a bit

  96. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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

  97. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yep

  98. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that's the total work required to assemble them.

  99. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok?

  100. woleraymond
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we have to add them up algebraically i guess?

  101. Algebraic!
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah

  102. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.