charges
two small charges each of mass m kg and charge Q C are suspended from a position by a sphere of L meter of negligible mass. if theta is the angle between vertical postiton to the equilibrium has been reached
prove: q2 = 4mgL2sin2(theata)tan(theata)k
the 2's are square
35 Replies
this question keeps haunting me
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.sphere of L METER?
A sphere?
I'm confuzzled
You mean strings?
yeah
why odes it haunt you
the question is weird idk
ehh just nlm
i cant find the way
Let tension be T
T cos theta = mg
T sin theta = electric force
its from charges
electric charges
hmm theres no G so no mututal gravitation
Eliminate T from these equations and you get your result
but howw
The sin²θ appears because horizontal distance of separation is 2lsinθ
Divide them na
tan theta aayega
Yup
Thanks Lance
dhyanwd
psr aise question me pata kaise chlega ki ky karna hai
what is your process?
just basic NLM
electrostatic force is just a force
equilibrium ki condition satisfy karni hai
These type of questions mostly involve like NLM and eliminating T.
if you're good at vectors then you can do it
To get some relation in theta and stuff.
I'd suggest revising some vectors and NLM
mujhe to smj hi nhi aya ho kya rha hai
mai to abhi is topic se introduce hui hu
ooo
ok
gravitation me nahi tha ye?
okkk
aise nhi kiya
hmm
mostly g ke formula alag alag wale
ya rato walr
oh you're a neet asp
this approach of physics haunts me tbh
yep
equilibrium dikhe to net force object pe 0 karne ka try karo
oh okkk
+solved @SirLancelotDuLac @Opt @Tanish
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