electrostatic pressure
a soap bubble of radius r is given a charge q uniformly over its surface. as a result of which the radius of bubble increases. if surface tension of soap is T then find increase in radius of bubble
36 Replies
@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.let initial pressure inside bubble be Pin and electrostatic pressure be Pele and increment in radius be x
solving these two eqns should give my answer but it doesnt
Trying doing it with energy instead
Pele = sigma^2 / 2 epsilon
T(4πr²)+self energy= T(4π)(r+x)²+new self energy
That makes no sense.
Wait
That makes more sense now
initial charge = 0 so self energy of bubble = 0
finally it would be kq^2/2(r+x)
Ok, yeah, that could also work
it doesnt give energy though
Initial surface energy = final surface energy + self energy?
dont u think solving this would be more difficult
it just doesnt seem convincing enough to me
it might be true tho
Try it out ig
.
I don't have paper with me.
Not really.
Let me go look for my workbook
alrr
Nvm that's a cubic
also T(8pir^2 )
two surfaces
Oh right.
ok so whats wrong in my method
Idk, I'm not too familiar with pressure ngl.
Best wait for someone else
alrr
did u give adv btw?
Yes
We don't talk about it.
.cengage
lol alrr
bitsat now
?
IAT
Bio grind
oh
best of luck
Thanks
ill just wait for some1 else
Assuming isothermal process it is rather P.V that remains constant and not the pressure. (ig, even tho the question should have specified it :/)
Wait, isothermal is giving a biquadratic :sweaty:
damn
why is my process wrong tho
This assumes adiabatic process no?
Yeah I don't have a clue. Do ping me if you get the answer.
Ah. One thing I don't get tho. Shouldn't dP also account for the change in excess pressure due to changing R?
brother book ka soln toh mere paas bhi hai 🙏
which book