Timeline for Force between two charged rods?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
Commonmark migration
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| Mar 23, 2013 at 10:42 | history | edited | Emilio Pisanty | CC BY-SA 3.0 |
Minor LaTeX correction.
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| Mar 23, 2013 at 0:56 | comment | added | Siva | When the two rods are parallel, the system has maximum potential energy. That means the torwue must be very small. But like I said, the equilibrium is unstable. So the torque will start small and as it slowly pushes the rods out of equilibrium, it will gather in strength. Imagine a ball rollong off a curved mountaintop (say: inverted parabola to leading order in the taylor expansion). It will start off very slowly, with a small force because the mountain is almost horizontal at the top (near an extrema). As the slope (of the mountain ~ potential energy) increases, so will the force. | |
| Mar 22, 2013 at 22:00 | comment | added | jclancy | I don't understand why the torque should be small for small $\theta$. Shouldn't the opposite be true, i.e. the torque should drive the rods from the $\theta = 0$ equilibrium extremely quickly? However I agree the radial component of the force should be zero. | |
| Mar 22, 2013 at 19:45 | comment | added | Siva | It seems to me that the radial component of the forceshould be zero in some nice frame... maybe when you sit in the frame where the hinge is stationary, but I'm not quite convinced. | |
| Mar 22, 2013 at 19:42 | history | answered | Siva | CC BY-SA 3.0 |