anonymous
  • anonymous
derive an expression for the work done in putting the four charges together as shown below
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1348178251421:dw|
anonymous
  • anonymous
|dw:1348178242336:dw|
anonymous
  • anonymous
#1 is free...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
#2 is -qV where V is kq/r so the work is -kq^2/a
anonymous
  • anonymous
#3 is -kq^2/a + kq^2/(a*sqrt(2))
anonymous
  • anonymous
see where this is going?
anonymous
  • anonymous
im trying to figgy it
anonymous
  • anonymous
|dw:1348178828104:dw|
anonymous
  • 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
  • anonymous
integrate and you get (kqq)/r ..at r=infinity it equals zero and at r=a it's (kqq)/a of course
anonymous
  • 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
  • anonymous
@Algebraic! note the distance for charge 2 to come the center is \[\sqrt{2a}\]
anonymous
  • anonymous
a
anonymous
  • anonymous
nothing is coming to the 'center'
anonymous
  • 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
  • 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
  • anonymous
|dw:1348180508889:dw|
anonymous
  • anonymous
@woleraymond get it?
anonymous
  • anonymous
@RaphaelFilgueiras so the work is the same moving a negative charge to a positive as moving a positive to a positive?
anonymous
  • anonymous
I see.
anonymous
  • anonymous
@RaphaelFilgueiras and @Algebraic! u guys are good
anonymous
  • anonymous
dead wrong
anonymous
  • anonymous
which direction is F in k(q)(-q) /r^2 and which direction is F in k(q)(q) /r^2 ??
anonymous
  • anonymous
@RaphaelFilgueiras what is the condition to have such sequence that didnt involve all the charge acting on each other
anonymous
  • anonymous
the work is the energy of configuration
anonymous
  • anonymous
lol do the work intergal @RaphaelFilgueiras
anonymous
  • anonymous
if you get a negative energy is a spontaneous config
anonymous
  • anonymous
|dw:1348180925907:dw|
anonymous
  • anonymous
yeah, two opposite charges move together spontaneously...
anonymous
  • anonymous
no you have to put it together so you will do work
anonymous
  • anonymous
in fact, they DO work... miracle of miracles.... because they'll actually ACCELERATE towards each other... wow
anonymous
  • anonymous
two positives charges???will atract each other?
anonymous
  • anonymous
yep @RaphaelFilgueiras work done from + to - should be -ve and + to + be +ve...what ypu think
anonymous
  • anonymous
mother of god http://i.imgur.com/jPZTm.gif
anonymous
  • anonymous
my halliday must be wrong
anonymous
  • anonymous
you're doing it wrong
anonymous
  • anonymous
ok,for sure
anonymous
  • anonymous
dude, just do the work integral, check the sign, end of story.
anonymous
  • anonymous
my way is rigth too
anonymous
  • anonymous
different answers, however. How shall we reconcile this???
anonymous
  • anonymous
you're doing it for all LIKE charges... these aren't all like charges....
anonymous
  • anonymous
no you must put the signal in formula!!!!
anonymous
  • anonymous
q1=-q
anonymous
  • anonymous
lol
anonymous
  • anonymous
@Mikael
anonymous
  • anonymous
now you're in trouble...
anonymous
  • anonymous
where you from @RaphaelFilgueiras ?
anonymous
  • anonymous
brazil?
anonymous
  • anonymous
yes
anonymous
  • anonymous
is your Halliday in Portuguese?
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
yep
anonymous
  • anonymous
now tell me +q * -q = +q *+q
anonymous
  • anonymous
better yet, find the page in Halliday that says that...
anonymous
  • anonymous
inb4 I couldn't find it but I know I saw it once
anonymous
  • anonymous
q1=-q;q2=+q;q3=-q;q4=+q
anonymous
  • anonymous
you got that much right... continue...
anonymous
  • 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
  • anonymous
or opposite.
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
you're conceding I take it?
anonymous
  • anonymous
nope,this was my first answer
anonymous
  • anonymous
what is yours?
anonymous
  • anonymous
omg you think that that page you linked is evidence IN your favor??
anonymous
  • anonymous
your answer is all positive terms...
anonymous
  • anonymous
no i put q1.q2
anonymous
  • anonymous
and said q1=-q ;q2=q....
anonymous
  • anonymous
there are no q1's or q2's on the sketch
anonymous
  • anonymous
there's q and -q
anonymous
  • anonymous
and you've already said you think the potential is positive independent of charge...
anonymous
  • anonymous
ok bye
anonymous
  • anonymous
bye.
anonymous
  • anonymous
@Algebraic! @RaphaelFilgueiras please what is the way forward on this
anonymous
  • anonymous
I did 3/4 terms of the solution, do you understand how we get them?
anonymous
  • anonymous
all that's left is to do the last charge...
anonymous
  • anonymous
hello?
anonymous
  • anonymous
RaphaelFilgueiras ended up doing it the correct way finally, so he gave you the answer...
anonymous
  • anonymous
are you there @woleraymond ?
anonymous
  • anonymous
yep im here @Algebraic! so tell me whats the final formula for this cos i dont get what @RaphaelFilgueiras did
anonymous
  • 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
  • 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
  • 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
  • anonymous
make sense?
anonymous
  • anonymous
i understand the integration but the second charge is moved from infinity to a distance a from the 3rd charge also
anonymous
  • 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
  • anonymous
brb equation editor is broken
anonymous
  • anonymous
this site is buggy
anonymous
  • 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
  • anonymous
and so on
anonymous
  • 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
  • anonymous
got it?
anonymous
  • 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
  • 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
  • anonymous
@woleraymond hello?
anonymous
  • anonymous
u making sense...@Algebraic! i got the import..will put every thing together in a bit
anonymous
  • 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
  • anonymous
yep
anonymous
  • anonymous
that's the total work required to assemble them.
anonymous
  • anonymous
ok?
anonymous
  • anonymous
we have to add them up algebraically i guess?
anonymous
  • anonymous
yeah

Looking for something else?

Not the answer you are looking for? Search for more explanations.