56 Replies
@Gyro Gearloose
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is my ans even dimensionally correct š
ans A hai
nope
sorry B
not A
yep
i missed 2 in denom
use work energy theorem
final potential energy is zero
initial potential energy calculate karlo
ezpz
potential energy kaise likhi
potential due to straight wire ka formula
then PE=qv
hmm nice idea
but why doesnt my method work
yeah im trying to solve using your method
it'll be a bad integration tho
nah i just subbed 4a-acostheta = t
accha accha
wait a min thats undefined for infinite wire
i'll take a look
nah just integrate 1/x
infinite ke lie potential diff defined hai
yes
okay so one thing i noticed
you forgot to take minus sign for negative charge
omg
wow.
omg major blunders in integration too
e
gg
damn
idea shi hai calc bohot galat
kisi ko aata ho toh bata do
yeah but tooooo much work bro šæ
only correct way up until now
wdym
aur koi approach hai ?
.
final toh 0 hai
potential difference is defined for infinite wire
initial is V due to inf long wire for one charge= integration of Edr= integral of (2klambda/r) dr
final potential diff is zero
yes cuz +q and -q at eaqual distances
and initial potential diff is Vb- Va = 2klambda ln3
iss mein dipole multiply kara for potential energy?
wdym multiply
seperate charges ka nikaalo
no kms typa name please
working bhejna apni
ok fair me likhar bhejta hu
give me back my original name pls š
potato lmaooo
š
@hardcoreisdead
u can also assume potential at infinity to be 0 which will lead to energy in 2nd case to be 0
alrr this makes sense
thanx
np
how to do using my method
Easy integral
Convert denom cos to sin, then take sin=t
Then convert to standard integral and solve
Assuming that integral is the right result
neeche cos^2 x hai
denominator comes out to be 3a-asin^2x = t
-2asinxcosx dx = dt
wait
Just take sinx=t
Not the whole thing
is this standard integral or smthing
Yup
But it won't give you a log
It'll give you a arctan
bro instead of using dipole treat +q and -q as individual charges
and then use work done=-deltaU
delta V for infinite wire is 2klamda ln(r2/r1)
$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} tan^{-1}(\frac{x}{a})$
Nimboi
yeah this
yeah aise aa gya
but whats wrong in my method
dipole lia hi nahi
are the ans equal?