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

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Algebraic!Best ResponseYou've already chosen the best response.2
dw:1348178251421:dw
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
dw:1348178242336:dw
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
#2 is qV where V is kq/r so the work is kq^2/a
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
#3 is kq^2/a + kq^2/(a*sqrt(2))
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
see where this is going?
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
im trying to figgy it
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
dw:1348178828104:dw
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.2
integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
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...
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.2
nothing is coming to the 'center'
 one year ago

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

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

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
dw:1348180508889:dw
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
@woleraymond get it?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
@RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
@RaphaelFilgueiras and @Algebraic! u guys are good
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
which direction is F in k(q)(q) /r^2 and which direction is F in k(q)(q) /r^2 ??
 one year ago

woleraymondBest 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
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
the work is the energy of configuration
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
lol do the work intergal @RaphaelFilgueiras
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
if you get a negative energy is a spontaneous config
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
dw:1348180925907:dw
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
yeah, two opposite charges move together spontaneously...
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
no you have to put it together so you will do work
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
two positives charges???will atract each other?
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
yep @RaphaelFilgueiras work done from + to  should be ve and + to + be +ve...what ypu think
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
mother of god http://i.imgur.com/jPZTm.gif
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
my halliday must be wrong
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
you're doing it wrong
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
dude, just do the work integral, check the sign, end of story.
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
my way is rigth too
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
different answers, however. How shall we reconcile this???
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
you're doing it for all LIKE charges... these aren't all like charges....
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
no you must put the signal in formula!!!!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
now you're in trouble...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
where you from @RaphaelFilgueiras ?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
is your Halliday in Portuguese?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
now tell me +q * q = +q *+q
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
better yet, find the page in Halliday that says that...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
inb4 I couldn't find it but I know I saw it once
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
q1=q;q2=+q;q3=q;q4=+q
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
you got that much right... continue...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
pretty much boils down to whether you think these are equal: \[\int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
you're conceding I take it?
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
nope,this was my first answer
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
what is yours?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
omg you think that that page you linked is evidence IN your favor??
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
your answer is all positive terms...
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
no i put q1.q2
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.0
and said q1=q ;q2=q....
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
there are no q1's or q2's on the sketch
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
and you've already said you think the potential is positive independent of charge...
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
@Algebraic! @RaphaelFilgueiras please what is the way forward on this
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
I did 3/4 terms of the solution, do you understand how we get them?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
all that's left is to do the last charge...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
are you there @woleraymond ?
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.2
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?
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.2
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
 one year ago

woleraymondBest 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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
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:
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
brb equation editor is broken
 one year ago

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

Algebraic!Best ResponseYou've already chosen the best response.2
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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
"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...
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
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?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
@woleraymond hello?
 one year ago

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

woleraymondBest 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
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
that's the total work required to assemble them.
 one year ago

woleraymondBest ResponseYou've already chosen the best response.0
we have to add them up algebraically i guess?
 one year ago
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