woleraymond
derive an expression for the work done in putting the four charges together as shown below
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Algebraic!
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|dw:1348178251421:dw|
woleraymond
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|dw:1348178242336:dw|
Algebraic!
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#1 is free...
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#2 is -qV where V is kq/r
so the work is -kq^2/a
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#3 is -kq^2/a + kq^2/(a*sqrt(2))
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see where this is going?
woleraymond
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im trying to figgy it
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|dw:1348178828104:dw|
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another way to look at it is :
\[W =\int\limits_{\infty}^{a} Fdr =\int\limits_{\infty}^{a} k \frac{ qq }{ r ^{2} } dr\]
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integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course
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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...
woleraymond
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@Algebraic! note the distance for charge 2 to come the center is \[\sqrt{2a}\]
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a
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nothing is coming to the 'center'
RaphaelFilgueiras
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the work is the total energy to assembly this configuration so:
\[w=1/4.\pi.\epsilon0[q1q2/a+q1q3/\sqrt{2a}+q1.q4/a+q2q3/a+q2q4/\sqrt{2a}+q3q4/a]\]
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you're asked to assemble the charges into the config. shown... that means moving #2 from infinity to a against the potential of #1
RaphaelFilgueiras
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|dw:1348180508889:dw|
RaphaelFilgueiras
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@woleraymond get it?
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@RaphaelFilgueiras
so the work is the same moving a negative charge to a positive as moving a positive to a positive?
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I see.
woleraymond
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@RaphaelFilgueiras and @Algebraic! u guys are good
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dead wrong
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which direction is F in k(q)(-q) /r^2 and which direction is F in k(q)(q) /r^2 ??
woleraymond
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@RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other
RaphaelFilgueiras
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the work is the energy of configuration
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lol do the work intergal @RaphaelFilgueiras
RaphaelFilgueiras
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if you get a negative energy is a spontaneous config
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|dw:1348180925907:dw|
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yeah, two opposite charges move together spontaneously...
RaphaelFilgueiras
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no you have to put it together so you will do work
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in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow
RaphaelFilgueiras
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two positives charges???will atract each other?
woleraymond
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yep @RaphaelFilgueiras work done from + to - should be -ve and + to + be +ve...what ypu think
RaphaelFilgueiras
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my halliday must be wrong
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you're doing it wrong
RaphaelFilgueiras
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ok,for sure
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dude, just do the work integral, check the sign, end of story.
RaphaelFilgueiras
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my way is rigth too
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different answers, however. How shall we reconcile this???
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you're doing it for all LIKE charges... these aren't all like charges....
RaphaelFilgueiras
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no you must put the signal in formula!!!!
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@Mikael
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now you're in trouble...
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where you from @RaphaelFilgueiras ?
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brazil?
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is your Halliday in Portuguese?
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yep
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now tell me +q * -q = +q *+q
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better yet, find the page in Halliday that says that...
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inb4 I couldn't find it but I know I saw it once
RaphaelFilgueiras
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q1=-q;q2=+q;q3=-q;q4=+q
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you got that much right... continue...
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pretty much boils down to whether you think these are equal:
\[\int\limits_{\infty}^{R} \frac{ k(q)(-q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]
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or opposite.
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you're conceding I take it?
RaphaelFilgueiras
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nope,this was my first answer
RaphaelFilgueiras
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what is yours?
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omg you think that that page you linked is evidence IN your favor??
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your answer is all positive terms...
RaphaelFilgueiras
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no i put q1.q2
RaphaelFilgueiras
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and said q1=-q ;q2=q....
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there are no q1's or q2's on the sketch
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there's q and -q
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and you've already said you think the potential is positive independent of charge...
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bye.
woleraymond
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@Algebraic! @RaphaelFilgueiras please what is the way forward on this
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I did 3/4 terms of the solution, do you understand how we get them?
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all that's left is to do the last charge...
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hello?
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RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...
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are you there @woleraymond ?
woleraymond
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yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did
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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?
woleraymond
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yes i agree, it takes no work to place the first charge as it must be -ve same also as the 3rd charge
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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
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make sense?
woleraymond
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i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also
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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:
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brb equation editor is broken
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this site is buggy
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\[W = \int\limits_{\infty}^{a} q \frac{ k(-q) }{ r ^{2} } dr +\int\limits_{\infty}^{a \sqrt{2}} -q \frac{ k(-q) }{ r ^{2}}dr\]
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and so on
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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
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got it?
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"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...
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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?
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@woleraymond hello?
woleraymond
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u making sense...@Algebraic! i got the import..will put every thing together in a bit
woleraymond
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@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
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yep
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that's the total work required to assemble them.
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ok?
woleraymond
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we have to add them up algebraically i guess?
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yeah