Silly physics...

Jochen Schroeder jschrod at
Thu Feb 13 01:54:42 PST 2003

Am Don, 2003-02-13 um 07.22 schrieb Rob Park:
> Alas! Jochen Schröder spake thus:
> > >Ah, you just gotta love physics...
> > 
> > Sorry but no! You have to multiply by k to get the error in F. Now if 
> > you want to calculate the error in r=distance between balls, you have to 
> > divide by k again. So you end up with a much lower error.
> Sorry, that's bullshit. If your force is so uncertain, how can you
> possibly get a certain distance out of it?
> You can't get a result that is more certain than it's component
> measurements. In other words, uncertainty always increases ;)
No, if you multiply by a constant your error stays the same, it might be
a higher number, but essentially it does not change. But an error in F
of lets say 10000 N might correspond to an error in distance of just
0.0000001 m it's still the same uncertainty
To take your example 
Coulomb-force is F=k(q1*q2)/r^2 where k=1/(4*pi*epsilon0) now lets
consider we don't have a pendulum but a spring this force corresponds to
a force F=Dz where D is the spring tension (I think this is the english
name for it) so k(q1*q2)/r^2=Dz now to to get z you have to divide by D
so we have (k/D)*q1*q2)/r^2=z now  this was my point. The way I said it
in my earlier post was: suppose you have F and q1, q2 given then
r=sqrt(F/(q1*q2*k)) note the k is in the denominator, it will be in the
error calculation as well.
What I don't quite understand in your example is you measured the
distance and wanted to calculate the charge of the balls. How could you
estimate the error for q without error-propagation. It seems to me you
did the whole calculation the wrong way around.

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