## anonymous 4 years ago derive an expression for the work done in putting the four charges together as shown below

1. anonymous

|dw:1348178251421:dw|

2. anonymous

|dw:1348178242336:dw|

3. anonymous

#1 is free...

4. anonymous

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

5. anonymous

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

6. anonymous

see where this is going?

7. anonymous

im trying to figgy it

8. anonymous

|dw:1348178828104:dw|

9. anonymous

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

10. anonymous

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

11. anonymous

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. anonymous

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

13. anonymous

a

14. anonymous

nothing is coming to the 'center'

15. anonymous

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. anonymous

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. anonymous

|dw:1348180508889:dw|

18. anonymous

@woleraymond get it?

19. anonymous

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

20. anonymous

I see.

21. anonymous

@RaphaelFilgueiras and @Algebraic! u guys are good

22. anonymous

23. anonymous

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

24. anonymous

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

25. anonymous

the work is the energy of configuration

26. anonymous

lol do the work intergal @RaphaelFilgueiras

27. anonymous

if you get a negative energy is a spontaneous config

28. anonymous

|dw:1348180925907:dw|

29. anonymous

yeah, two opposite charges move together spontaneously...

30. anonymous

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

31. anonymous

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

32. anonymous

two positives charges???will atract each other?

33. anonymous

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

34. anonymous

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

35. anonymous

my halliday must be wrong

36. anonymous

you're doing it wrong

37. anonymous

ok,for sure

38. anonymous

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

39. anonymous

my way is rigth too

40. anonymous

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

41. anonymous

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

42. anonymous

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

43. anonymous

q1=-q

44. anonymous

lol

45. anonymous

@Mikael

46. anonymous

now you're in trouble...

47. anonymous

where you from @RaphaelFilgueiras ?

48. anonymous

brazil?

49. anonymous

yes

50. anonymous

51. anonymous

52. anonymous

yep

53. anonymous

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

54. anonymous

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

55. anonymous

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

56. anonymous

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

57. anonymous

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

58. anonymous

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. anonymous

or opposite.

60. anonymous

61. anonymous

you're conceding I take it?

62. anonymous

63. anonymous

what is yours?

64. anonymous

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

65. anonymous

66. anonymous

no i put q1.q2

67. anonymous

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

68. anonymous

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

69. anonymous

there's q and -q

70. anonymous

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

71. anonymous

ok bye

72. anonymous

bye.

73. anonymous

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

74. anonymous

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

75. anonymous

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

76. anonymous

hello?

77. anonymous

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

78. anonymous

are you there @woleraymond ?

79. anonymous

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

80. anonymous

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. anonymous

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

82. anonymous

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. anonymous

make sense?

84. anonymous

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

85. anonymous

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. anonymous

brb equation editor is broken

87. anonymous

this site is buggy

88. anonymous

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

89. anonymous

and so on

90. anonymous

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. anonymous

got it?

92. anonymous

"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. anonymous

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. anonymous

@woleraymond hello?

95. anonymous

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

96. anonymous

@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. anonymous

yep

98. anonymous

that's the total work required to assemble them.

99. anonymous

ok?

100. anonymous

we have to add them up algebraically i guess?

101. anonymous

yeah