derive an expression for the work done in putting the four charges together as shown below

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derive an expression for the work done in putting the four charges together as shown below

Physics
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|dw:1348178251421:dw|
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#1 is free...

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Other answers:

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

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