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|dw:1348178251421:dw|

|dw:1348178242336:dw|

#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))

see where this is going?

im trying to figgy it

|dw:1348178828104:dw|

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

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

nothing is coming to the 'center'

|dw:1348180508889:dw|

@woleraymond get it?

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 ??

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

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?

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...

or opposite.

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 ?

make sense?

brb equation editor is broken

this site is buggy

and so on

got it?

@woleraymond hello?

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

yep

that's the total work required to assemble them.

ok?

we have to add them up algebraically i guess?

yeah