Recent content by Curieuse

Thanks very much sir! :D

Ah so I have been confusing angles and magnitudes.. τapp.dθ has to be positive yes! :O U(θ2)-U(θ1)=-pE(cos θ2-cosθ1) so if we take U(π/2)=0, then U(0)=-pE. Is this correct?

since τfield= p x E is intrinsically negative(?) i put the negative sign in front of it to make it positive in τapp..

Then i'd be getting the work done by the field right? I am trying to get work done by external agent in turning the dipole as in the figure in my previous post.. The turning i indicated was that produced by external torque..this work is stored in the system right?

τapp is -p x E will mean that it points out of the plane in this figure right? But how can that be? How do I derive potential energy in this scenario? the violet arrow is the dipole moment vector p. :cry: I know I am missing something big but i am not able to figure out why..

But isn't it only in the work integral that it becomes a - wrt +dθ? because Wfield=∫τfielddθcos π?

Homework Statement To derive Potential Energy for dipole p in Electric Field E. 2. Homework Equations Potential Energy is the work done by the external agent in turning the angle of the dipole from the U=0 position to another position against the influence of the electric field applied...

:oldsurprised: Decode :oldbiggrin: My bad!

No but it wasn't that laborious, Newtons method! Just one tricky fraction at the end! o_O

Yes i should've done that really! But being blinded by all the algebra i failed to see it was the decimal which was to just be limited at some point! :P I learned the method explained in this page http://www.basic-mathematics.com/square-root-algorithm.html I have no idea why it works, but I'll...

Yep! I tried this too! Started with 2/1. By x4 i got 7e(-7) error! When does this method not work? Like for non differentiable fns it won't, of course! Any other things one should keep in mind when applying this?

Thanks ray! This method rocks totally! I'm so happy i learned this! It's ingenious!:oldbiggrin: :oldlaugh:

Is it by trial and error that we get the fraction that we write the continued fraction for? I tried with 2.6457514 and got 2+1/1+1/1+1/1+1/4+1/1+1/1+1/1+1/4+1/1+/1+1/1+1/4+1/1+87791/159294 I realized the quotients were repeating and this stopped. I should solve for each and get the series 2, 3...

I now see why you took denominator squared to be less than the allowed error! It comes from continuous fractions, if I am not wrong. Also the fraction is simply to large to work with. As Scientific 601 said, the error is within limits only in the 13th term. Is there any way of doing this faster...

Well, thanks for this! I was blinded by all the algebra! It's so simple! T.T But i still want to know the algebraic way of doing this! Why is q > 10^6 ? Are we just choosing an arbitrary denominator and arriving at the numerator from applying the conditions? I should be missing something here!