A water drop in electrostatic orbit around a needle in 0𝑔pic.twitter.com/G2x3Ud195K
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What's going on? • The needle has been given an electrostatic charge by rubbing it on silk • Water droplets are placed by the needle w/ velocity v≠ 0 • H₂O has a slight separation of charge (i.e a dipole moment) so it feels the needle's charge
+ Coulomb's law --> orbit4 replies 11 retweets 135 likesShow this thread -
Replying to @InertialObservr
Are you sure this is correct? The force due to induced polarization is ~ 1/r^5 for a point charge, and ~ 1/r^3 for a line charge, neither of which admits a stable orbit.
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Replying to @gravity_levity @InertialObservr
Just to be pedantic (excuse me) the normal Coulomb force admits a stable circular orbit because the attractive force is ~ 1/r^2, while the outward "centrifugal force" is mv^2/r = L^2/(m r^3).
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Replying to @gravity_levity @InertialObservr
So if r is perturbed slightly from its stable equilibrium the orbit is either pushed back out (by the larger centrifugal force) or back in (by the larger Coulomb force).
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Replying to @gravity_levity @InertialObservr
For a dielectric sphere attracted to a point charge, the force is F = d(p E)/dr, where p ~ R^3 E is the induced dipole moment (R is the radius of the sphere). So F ~ 1/r^5, and this does not admit a stable orbit. For a dielectric sphere attracted to a line charge F ~ 1/r^3
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Replying to @gravity_levity @InertialObservr
F ~ 1/r^3 does not admit a truly stable orbit, although maybe you could get a few periods of marginal stability.
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Replying to @gravity_levity
is there a charge distribution of the needle that could give a stable orbit?
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Replying to @InertialObservr
That's a good question, but I suspect that the answer is no, since both extreme cases (point charge and uniform line charge) are unstable. Of course, if the water drop has a net charge then you immediately get stability.
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hmm.. the motion to me doesn't look totally stable and i don't know the natural trajectories of the 1/r^3 1/r^5 potentials
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