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woleraymond
Group Title
derive an expression for the work done in putting the four charges together as shown below
 2 years ago
 2 years ago
woleraymond Group Title
derive an expression for the work done in putting the four charges together as shown below
 2 years ago
 2 years ago

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Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
dw:1348178251421:dw
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
dw:1348178242336:dw
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
#1 is free...
 2 years ago

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

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
see where this is going?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
im trying to figgy it
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
dw:1348178828104:dw
 2 years ago

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

Algebraic! Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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...
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
nothing is coming to the 'center'
 2 years ago

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

Algebraic! Group TitleBest 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
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
dw:1348180508889:dw
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
@woleraymond get it?
 2 years ago

Algebraic! Group TitleBest 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?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@RaphaelFilgueiras and @Algebraic! u guys are good
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
dead wrong
 2 years ago

Algebraic! Group TitleBest 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 ??
 2 years ago

woleraymond Group TitleBest 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
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
the work is the energy of configuration
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
lol do the work intergal @RaphaelFilgueiras
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
if you get a negative energy is a spontaneous config
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
dw:1348180925907:dw
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
yeah, two opposite charges move together spontaneously...
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
no you have to put it together so you will do work
 2 years ago

Algebraic! Group TitleBest 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
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
two positives charges???will atract each other?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
yep @RaphaelFilgueiras work done from + to  should be ve and + to + be +ve...what ypu think
 2 years ago

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

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
my halliday must be wrong
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
you're doing it wrong
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
ok,for sure
 2 years ago

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

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
my way is rigth too
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
different answers, however. How shall we reconcile this???
 2 years ago

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

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
no you must put the signal in formula!!!!
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
q1=q
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
lol
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
now you're in trouble...
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
where you from @RaphaelFilgueiras ?
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
yes
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
is your Halliday in Portuguese?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
now tell me +q * q = +q *+q
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
better yet, find the page in Halliday that says that...
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
inb4 I couldn't find it but I know I saw it once
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
q1=q;q2=+q;q3=q;q4=+q
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
you got that much right... continue...
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
or opposite.
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
you're conceding I take it?
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
nope,this was my first answer
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
what is yours?
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
your answer is all positive terms...
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
no i put q1.q2
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
and said q1=q ;q2=q....
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
there are no q1's or q2's on the sketch
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
there's q and q
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
and you've already said you think the potential is positive independent of charge...
 2 years ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
ok bye
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
@Algebraic! @RaphaelFilgueiras please what is the way forward on this
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
all that's left is to do the last charge...
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
are you there @woleraymond ?
 2 years ago

woleraymond Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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?
 2 years ago

woleraymond Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
make sense?
 2 years ago

woleraymond Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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:
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
brb equation editor is broken
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
this site is buggy
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
and so on
 2 years ago

Algebraic! Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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...
 2 years ago

Algebraic! Group TitleBest 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?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
@woleraymond hello?
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
u making sense...@Algebraic! i got the import..will put every thing together in a bit
 2 years ago

woleraymond Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
that's the total work required to assemble them.
 2 years ago

woleraymond Group TitleBest ResponseYou've already chosen the best response.0
we have to add them up algebraically i guess?
 2 years ago
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