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

- anonymous

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

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- anonymous

|dw:1348178251421:dw|

- anonymous

|dw:1348178242336:dw|

- anonymous

#1 is free...

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## More answers

- anonymous

#2 is -qV where V is kq/r so the work is -kq^2/a

- anonymous

#3 is -kq^2/a + kq^2/(a*sqrt(2))

- anonymous

see where this is going?

- anonymous

im trying to figgy it

- anonymous

|dw:1348178828104:dw|

- anonymous

another way to look at it is : \[W =\int\limits_{\infty}^{a} Fdr =\int\limits_{\infty}^{a} k \frac{ qq }{ r ^{2} } dr\]

- anonymous

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

- anonymous

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...

- anonymous

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

- anonymous

a

- anonymous

nothing is coming to the 'center'

- anonymous

the work is the total energy to assembly this configuration so: \[w=1/4.\pi.\epsilon0[q1q2/a+q1q3/\sqrt{2a}+q1.q4/a+q2q3/a+q2q4/\sqrt{2a}+q3q4/a]\]

- anonymous

you're asked to assemble the charges into the config. shown... that means moving #2 from infinity to a against the potential of #1

- anonymous

|dw:1348180508889:dw|

- anonymous

@woleraymond get it?

- anonymous

@RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?

- anonymous

I see.

- anonymous

@RaphaelFilgueiras and @Algebraic! u guys are good

- anonymous

dead wrong

- anonymous

which direction is F in k(q)(-q) /r^2 and which direction is F in k(q)(q) /r^2 ??

- anonymous

@RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other

- anonymous

the work is the energy of configuration

- anonymous

lol do the work intergal @RaphaelFilgueiras

- anonymous

if you get a negative energy is a spontaneous config

- anonymous

|dw:1348180925907:dw|

- anonymous

yeah, two opposite charges move together spontaneously...

- anonymous

no you have to put it together so you will do work

- anonymous

in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow

- anonymous

two positives charges???will atract each other?

- anonymous

yep @RaphaelFilgueiras work done from + to - should be -ve and + to + be +ve...what ypu think

- anonymous

mother of god http://i.imgur.com/jPZTm.gif

- anonymous

my halliday must be wrong

- anonymous

you're doing it wrong

- anonymous

ok,for sure

- anonymous

dude, just do the work integral, check the sign, end of story.

- anonymous

my way is rigth too

- anonymous

different answers, however. How shall we reconcile this???

- anonymous

you're doing it for all LIKE charges... these aren't all like charges....

- anonymous

no you must put the signal in formula!!!!

- anonymous

q1=-q

- anonymous

lol

- anonymous

@Mikael

- anonymous

now you're in trouble...

- anonymous

where you from @RaphaelFilgueiras ?

- anonymous

brazil?

- anonymous

yes

- anonymous

is your Halliday in Portuguese?

- anonymous

##### 1 Attachment

- anonymous

yep

- anonymous

now tell me +q * -q = +q *+q

- anonymous

better yet, find the page in Halliday that says that...

- anonymous

inb4 I couldn't find it but I know I saw it once

- anonymous

q1=-q;q2=+q;q3=-q;q4=+q

- anonymous

you got that much right... continue...

- anonymous

pretty much boils down to whether you think these are equal: \[\int\limits_{\infty}^{R} \frac{ k(q)(-q) }{ r } dr = \int\limits_{\infty}^{R} \frac{ k(q)(q) }{ r } dr\]

- anonymous

or opposite.

- anonymous

##### 1 Attachment

- anonymous

you're conceding I take it?

- anonymous

nope,this was my first answer

- anonymous

what is yours?

- anonymous

omg you think that that page you linked is evidence IN your favor??

- anonymous

your answer is all positive terms...

- anonymous

no i put q1.q2

- anonymous

and said q1=-q ;q2=q....

- anonymous

there are no q1's or q2's on the sketch

- anonymous

there's q and -q

- anonymous

and you've already said you think the potential is positive independent of charge...

- anonymous

ok bye

- anonymous

bye.

- anonymous

@Algebraic! @RaphaelFilgueiras please what is the way forward on this

- anonymous

I did 3/4 terms of the solution, do you understand how we get them?

- anonymous

all that's left is to do the last charge...

- anonymous

hello?

- anonymous

RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...

- anonymous

are you there @woleraymond ?

- anonymous

yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did

- anonymous

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?

- anonymous

yes i agree, it takes no work to place the first charge as it must be -ve same also as the 3rd charge

- anonymous

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

- anonymous

make sense?

- anonymous

i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also

- anonymous

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:

- anonymous

brb equation editor is broken

- anonymous

this site is buggy

- anonymous

\[W = \int\limits_{\infty}^{a} q \frac{ k(-q) }{ r ^{2} } dr +\int\limits_{\infty}^{a \sqrt{2}} -q \frac{ k(-q) }{ r ^{2}}dr\]

- anonymous

and so on

- anonymous

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

- anonymous

got it?

- anonymous

"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

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?

- anonymous

@woleraymond hello?

- anonymous

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

- anonymous

@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

yep

- anonymous

that's the total work required to assemble them.

- anonymous

ok?

- anonymous

we have to add them up algebraically i guess?

- anonymous

yeah

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