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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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2. woleraymond Group Title

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3. Algebraic! Group Title

#1 is free...

4. Algebraic! Group Title

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

5. Algebraic! Group Title

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

6. Algebraic! Group Title

see where this is going?

7. woleraymond Group Title

im trying to figgy it

8. Algebraic! Group Title

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9. Algebraic! Group Title

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! Group Title

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

11. Algebraic! Group Title

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 Group Title

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

13. Algebraic! Group Title

a

14. Algebraic! Group Title

nothing is coming to the 'center'

15. RaphaelFilgueiras Group Title

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! Group Title

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 Group Title

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18. RaphaelFilgueiras Group Title

@woleraymond get it?

19. Algebraic! Group Title

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

20. Algebraic! Group Title

I see.

21. woleraymond Group Title

@RaphaelFilgueiras and @Algebraic! u guys are good

22. Algebraic! Group Title

dead wrong

23. Algebraic! Group Title

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

24. woleraymond Group Title

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

25. RaphaelFilgueiras Group Title

the work is the energy of configuration

26. Algebraic! Group Title

lol do the work intergal @RaphaelFilgueiras

27. RaphaelFilgueiras Group Title

if you get a negative energy is a spontaneous config

28. Algebraic! Group Title

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29. Algebraic! Group Title

yeah, two opposite charges move together spontaneously...

30. RaphaelFilgueiras Group Title

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

31. Algebraic! Group Title

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

32. RaphaelFilgueiras Group Title

two positives charges???will atract each other?

33. woleraymond Group Title

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

34. Algebraic! Group Title

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

35. RaphaelFilgueiras Group Title

my halliday must be wrong

36. Algebraic! Group Title

you're doing it wrong

37. RaphaelFilgueiras Group Title

ok,for sure

38. Algebraic! Group Title

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

39. RaphaelFilgueiras Group Title

my way is rigth too

40. Algebraic! Group Title

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

41. Algebraic! Group Title

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

42. RaphaelFilgueiras Group Title

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

43. RaphaelFilgueiras Group Title

q1=-q

44. RaphaelFilgueiras Group Title

lol

45. Algebraic! Group Title

@Mikael

46. Algebraic! Group Title

now you're in trouble...

47. Algebraic! Group Title

where you from @RaphaelFilgueiras ?

48. Algebraic! Group Title

brazil?

49. RaphaelFilgueiras Group Title

yes

50. Algebraic! Group Title

is your Halliday in Portuguese?

51. RaphaelFilgueiras Group Title

52. Algebraic! Group Title

yep

53. Algebraic! Group Title

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

54. Algebraic! Group Title

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

55. Algebraic! Group Title

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

56. RaphaelFilgueiras Group Title

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

57. Algebraic! Group Title

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

58. Algebraic! Group Title

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! Group Title

or opposite.

60. RaphaelFilgueiras Group Title

61. Algebraic! Group Title

you're conceding I take it?

62. RaphaelFilgueiras Group Title

nope,this was my first answer

63. RaphaelFilgueiras Group Title

what is yours?

64. Algebraic! Group Title

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

65. Algebraic! Group Title

your answer is all positive terms...

66. RaphaelFilgueiras Group Title

no i put q1.q2

67. RaphaelFilgueiras Group Title

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

68. Algebraic! Group Title

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

69. Algebraic! Group Title

there's q and -q

70. Algebraic! Group Title

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

71. RaphaelFilgueiras Group Title

ok bye

72. Algebraic! Group Title

bye.

73. woleraymond Group Title

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

74. Algebraic! Group Title

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

75. Algebraic! Group Title

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

76. Algebraic! Group Title

hello?

77. Algebraic! Group Title

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

78. Algebraic! Group Title

are you there @woleraymond ?

79. woleraymond Group Title

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

80. Algebraic! Group Title

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 Group Title

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! Group Title

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! Group Title

make sense?

84. woleraymond Group Title

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

85. Algebraic! Group Title

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! Group Title

brb equation editor is broken

87. Algebraic! Group Title

this site is buggy

88. Algebraic! Group Title

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

89. Algebraic! Group Title

and so on

90. Algebraic! Group Title

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! Group Title

got it?

92. Algebraic! Group Title

"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! Group Title

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! Group Title

@woleraymond hello?

95. woleraymond Group Title

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

96. woleraymond Group Title

@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! Group Title

yep

98. Algebraic! Group Title

that's the total work required to assemble them.

99. Algebraic! Group Title

ok?

100. woleraymond Group Title

we have to add them up algebraically i guess?

101. Algebraic! Group Title

yeah