## woleraymond derive an expression for the work done in putting the four charges together as shown below one year ago one year ago

1. Algebraic!

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2. woleraymond

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3. Algebraic!

#1 is free...

4. Algebraic!

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

5. Algebraic!

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

6. Algebraic!

see where this is going?

7. woleraymond

im trying to figgy it

8. Algebraic!

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9. Algebraic!

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!

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

11. Algebraic!

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

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

13. Algebraic!

a

14. Algebraic!

nothing is coming to the 'center'

15. RaphaelFilgueiras

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!

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

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18. RaphaelFilgueiras

@woleraymond get it?

19. Algebraic!

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

20. Algebraic!

I see.

21. woleraymond

@RaphaelFilgueiras and @Algebraic! u guys are good

22. Algebraic!

23. Algebraic!

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

24. woleraymond

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

25. RaphaelFilgueiras

the work is the energy of configuration

26. Algebraic!

lol do the work intergal @RaphaelFilgueiras

27. RaphaelFilgueiras

if you get a negative energy is a spontaneous config

28. Algebraic!

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29. Algebraic!

yeah, two opposite charges move together spontaneously...

30. RaphaelFilgueiras

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

31. Algebraic!

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

32. RaphaelFilgueiras

two positives charges???will atract each other?

33. woleraymond

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

34. Algebraic!

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

35. RaphaelFilgueiras

my halliday must be wrong

36. Algebraic!

you're doing it wrong

37. RaphaelFilgueiras

ok,for sure

38. Algebraic!

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

39. RaphaelFilgueiras

my way is rigth too

40. Algebraic!

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

41. Algebraic!

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

42. RaphaelFilgueiras

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

43. RaphaelFilgueiras

q1=-q

44. RaphaelFilgueiras

lol

45. Algebraic!

@Mikael

46. Algebraic!

now you're in trouble...

47. Algebraic!

where you from @RaphaelFilgueiras ?

48. Algebraic!

brazil?

49. RaphaelFilgueiras

yes

50. Algebraic!

51. RaphaelFilgueiras

52. Algebraic!

yep

53. Algebraic!

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

54. Algebraic!

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

55. Algebraic!

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

56. RaphaelFilgueiras

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

57. Algebraic!

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

58. Algebraic!

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!

or opposite.

60. RaphaelFilgueiras

61. Algebraic!

you're conceding I take it?

62. RaphaelFilgueiras

63. RaphaelFilgueiras

what is yours?

64. Algebraic!

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

65. Algebraic!

66. RaphaelFilgueiras

no i put q1.q2

67. RaphaelFilgueiras

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

68. Algebraic!

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

69. Algebraic!

there's q and -q

70. Algebraic!

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

71. RaphaelFilgueiras

ok bye

72. Algebraic!

bye.

73. woleraymond

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

74. Algebraic!

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

75. Algebraic!

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

76. Algebraic!

hello?

77. Algebraic!

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

78. Algebraic!

are you there @woleraymond ?

79. woleraymond

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

80. Algebraic!

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

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!

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!

make sense?

84. woleraymond

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

85. Algebraic!

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!

brb equation editor is broken

87. Algebraic!

this site is buggy

88. Algebraic!

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

89. Algebraic!

and so on

90. Algebraic!

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!

got it?

92. Algebraic!

"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!

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!

@woleraymond hello?

95. woleraymond

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

96. woleraymond

@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!

yep

98. Algebraic!

that's the total work required to assemble them.

99. Algebraic!

ok?

100. woleraymond

we have to add them up algebraically i guess?

101. Algebraic!

yeah