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Author Topic: F = KeQ1Q2/r2 ?  (Read 4639 times)

Ron Hughes

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F = KeQ1Q2/r2 ?
« on: 04/04/2011 23:50:38 »
The equation, F = KeQ1Q2/r2 as everyone knows, is used to determine the force between two charged particles at some distance r. Suppose Q2 was racing around inside a donut shaped tube, would that change the force on Q1?

Mod edit: I included the equation in the post title to attract the appropriate attention !
« Last Edit: 05/04/2011 01:07:32 by neilep »

Phractality

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F = KeQ1Q2/r2 ?
« Reply #1 on: 05/04/2011 02:04:22 »
There would be a magnetic force in addition to the electrostatic force, and the distance, r, would be variable unless the stationary charge is on the donuts axis. At relativistic speed, you would have to apply GR, since the problem involves acceleration. In fact, I think the magnetic force can be derived by applying GR to the electrostatic forces.

Geezer

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F = KeQ1Q2/r2 ?
« Reply #2 on: 05/04/2011 08:27:36 »

The equation, F = KeQ1Q2/r2 as everyone knows,

Well, I'm pretty sure Neil knows it, but I think I forgot it. I'm sure I'll get dinged for that.

imatfaal

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F = KeQ1Q2/r2 ?
« Reply #3 on: 05/04/2011 11:02:30 »
I would go along with Fract.  Remember Coulomb's law gives the ElectroSTATIC attraction or repulsion.  At slow speeds it can be approximated - but only when the curl of Electric field AND the time derivative of the magnetic field are damn close to zero

I am sure someone has done the heavyduty maths somewhere on the web to give the answer to your question

lightarrow

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F = KeQ1Q2/r2 ?
« Reply #4 on: 05/04/2011 21:03:11 »
The equation, F = KeQ1Q2/r2 as everyone knows, is used to determine the force between two charged particles at some distance r. Suppose Q2 was racing around inside a donut shaped tube, would that change the force on Q1?
Q1 is still in the lab frame? Where is it, in the doughnut' centre?

Ron Hughes

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F = KeQ1Q2/r2 ?
« Reply #5 on: 05/04/2011 21:24:12 »
It's running around the donut tube at 0.5 C.
« Last Edit: 05/04/2011 21:25:44 by Ron Hughes »

imatfaal

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F = KeQ1Q2/r2 ?
« Reply #6 on: 06/04/2011 16:00:11 »
at half the speed of light then coulombs law is no longer a valid approximation - your answer is the end of a bunch of nasty differential equations and tensor analysis

Ron Hughes

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F = KeQ1Q2/r2 ?
« Reply #7 on: 06/04/2011 17:34:50 »
Suppose the donut only has a diameter of 10-12nm

imatfaal

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F = KeQ1Q2/r2 ?
« Reply #8 on: 06/04/2011 18:10:06 »
At 10^-21m all bets are off when considering a particle at relativistic speed whizzing around a torus.  any particle would radiate energy away at a huge rate and drop to a more normal speed - even then we could not use coulomb's law

lightarrow

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F = KeQ1Q2/r2 ?
« Reply #9 on: 06/04/2011 19:45:57 »
It's running around the donut tube at 0.5 C.
Wait a moment: in your first post you wrote that Q2 is running around the doughnute; now you say that even Q1 is doing it?

Ron Hughes

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F = KeQ1Q2/r2 ?
« Reply #10 on: 07/04/2011 03:54:32 »
My apology, your right I meant Q2. Q1 is stationary.

lightarrow

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F = KeQ1Q2/r2 ?
« Reply #11 on: 07/04/2011 12:43:24 »
My apology, your right I meant Q2. Q1 is stationary.

Ok. Then, *classically* the force on Q1 doesn't change (I believe) because you don't have to change frame of reference and because Lorentz force on Q1 is zero.

Obviously, since you asked about a doughnut of 10-12 nm, the classical description is totally wrong. It's totally wrong even at 0.1 nm, that is, in the case of an hydrogen atom, so you can easily understand what value could have at 0.000000000001 nm...
« Last Edit: 07/04/2011 12:45:51 by lightarrow »

Ron Hughes

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F = KeQ1Q2/r2 ?
« Reply #12 on: 07/04/2011 15:15:32 »
With Q2 running around in a circle it has a time varying field. Wouldn't this produce a new force on Q1?

lightarrow

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F = KeQ1Q2/r2 ?
« Reply #13 on: 07/04/2011 20:50:17 »
With Q2 running around in a circle it has a time varying field. Wouldn't this produce a new force on Q1?
A time varying electric field produces a magnetic field, which, through Lorentz Force, produces a new force on a moving charge. But here Q1 is stationary, so there shouldn't be other forces (however I will check this).
The electric field generated from Q2 should be computed using the "retarded" formula, but in this case nothing varies, because of the cylindrical symmetry of the system.

Ron Hughes

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F = KeQ1Q2/r2 ?
« Reply #14 on: 09/04/2011 04:13:55 »
If Q1 is situated on the rim of the tube the field of Q2 will be stronger when near and weaker when on the opposite side.

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F = KeQ1Q2/r2 ?
« Reply #14 on: 09/04/2011 04:13:55 »