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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348178251421:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348178242336:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0#2 is qV where V is kq/r so the work is kq^2/a

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0#3 is kq^2/a + kq^2/(a*sqrt(2))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see where this is going?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im trying to figgy it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348178828104:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0another way to look at it is : \[W =\int\limits_{\infty}^{a} Fdr =\int\limits_{\infty}^{a} k \frac{ qq }{ r ^{2} } dr\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! note the distance for charge 2 to come the center is \[\sqrt{2a}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nothing is coming to the 'center'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the work is the total energy to assembly this configuration so: \[w=1/4.\pi.\epsilon0[q1q2/a+q1q3/\sqrt{2a}+q1.q4/a+q2q3/a+q2q4/\sqrt{2a}+q3q4/a]\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're asked to assemble the charges into the config. shown... that means moving #2 from infinity to a against the potential of #1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348180508889:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RaphaelFilgueiras and @Algebraic! u guys are good

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which direction is F in k(q)(q) /r^2 and which direction is F in k(q)(q) /r^2 ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the work is the energy of configuration

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol do the work intergal @RaphaelFilgueiras

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you get a negative energy is a spontaneous config

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348180925907:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, two opposite charges move together spontaneously...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no you have to put it together so you will do work

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0two positives charges???will atract each other?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep @RaphaelFilgueiras work done from + to  should be ve and + to + be +ve...what ypu think

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mother of god http://i.imgur.com/jPZTm.gif

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my halliday must be wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're doing it wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dude, just do the work integral, check the sign, end of story.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0different answers, however. How shall we reconcile this???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're doing it for all LIKE charges... these aren't all like charges....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no you must put the signal in formula!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now you're in trouble...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where you from @RaphaelFilgueiras ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is your Halliday in Portuguese?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now tell me +q * q = +q *+q

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0better yet, find the page in Halliday that says that...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0inb4 I couldn't find it but I know I saw it once

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0q1=q;q2=+q;q3=q;q4=+q

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you got that much right... continue...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pretty much boils down to whether you think these are equal: \[\int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're conceding I take it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nope,this was my first answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0omg you think that that page you linked is evidence IN your favor??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0your answer is all positive terms...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and said q1=q ;q2=q....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are no q1's or q2's on the sketch

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and you've already said you think the potential is positive independent of charge...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! @RaphaelFilgueiras please what is the way forward on this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did 3/4 terms of the solution, do you understand how we get them?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0all that's left is to do the last charge...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you there @woleraymond ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i agree, it takes no work to place the first charge as it must be ve same also as the 3rd charge

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0brb equation editor is broken

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[W = \int\limits_{\infty}^{a} q \frac{ k(q) }{ r ^{2} } dr +\int\limits_{\infty}^{a \sqrt{2}} q \frac{ k(q) }{ r ^{2}}dr\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0W 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0charge 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u making sense...@Algebraic! i got the import..will put every thing together in a bit

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the total work required to assemble them.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we have to add them up algebraically i guess?
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