# Perpetual motion machine [duplicate]

I'm having trouble understanding why this machine doesn't work. Part of a ring, half of which is uniformly charged, is located between two oppositely charged plates, attracting to the negative one and pushing away from the positive one, thus gaining angular momentum.

I understand that perpetuum mobile is impossible, however, I'm interested why in terms of electrostatics and mechanics this machine can't accelerate infinitely.

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## marked as duplicate by John Rennie, Sebastian Riese, John Duffield, ACuriousMind, Kyle KanosFeb 23 at 17:51

Possible duplicate of Can magnets rotate infinitely? – GPhys Feb 23 at 8:54
It doesn't work for the same reason as always: energy conservation. – CuriousOne Feb 23 at 9:41
Is the ring or the capacitor moving? – gonenc Feb 23 at 9:41
y-projection? Which direction is y? – Jens Feb 23 at 13:23
I think that the expected rotation direction is inverted in the picture? – user2464424 Feb 23 at 15:56

I have reconstructed your diagram computationally and numerically computed the potential energy as a function of the orientation of the ring:

As you can see, when we neglect friction (something that you can't do in the real world) the machine is indeed capable of perpetual motion. However, some important points:

1. The total change in energy of the system per revolution is zero, so if you attempt to extract energy (or friction acts) then the device will grind to a halt. This is independent of how large you make the E-field.
2. In order for the system to complete a revolution, you need to put in the requisite energy. Either by starting the system at the peak of the potential energy, or giving it sufficient kinetic energy to overcome the peaks. You might as well remove the charges entirely and just spin a wheel.
3. You may wonder whether dynamically adjusting the E-field in order to increase the acceleration-stage and attenuate the deceleration-stage would allow you to overcome friction/extract energy; if you do the calculation you will find that it takes as much energy to dynamically adjust the E-field as you get out of it.
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"you will find that it takes as much energy to dynamically adjust the E-field as you get out of it." - minus EM radiation. The amount may be miniscule but it's unavoidable. – Jan Dvorak Feb 23 at 15:51
I am slightly confused. The ring is uniformly charged, thus its motion is at most flat, without accelerating nor decelerating no matter its orientation. How did you come up with that picture? – user2464424 Feb 23 at 16:01
@user2464424 Only half of the ring is charged (to give it a kick). If the entire ring were charged then the PE (and velocity) would indeed be constant. – lemon Feb 23 at 16:07

Perhaps a gravitational analogy will help?
The positive charges are railway trucks on a circular track in a frictionless environment. The positive plate is the top of a hill and the negative plate is the bottom of the hill.

Set the railway trucks moving
The trucks going down the hill lose gravitational potential energy whilst there are trucks going up the hill gaining the same amount of gravitational energy.

So in theory the trucks can remain in motion for ever but no energy can be extracted because that will slow down or stop the trucks.
And there is also friction to consider.

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Thanks, but, as I said, I can't understand, which mechanisms exactly prevent the ring from accelerating. Even if we consider friction, E-field between the plates can be made large enough to overcome it. – Dmst Feb 23 at 9:28
You have missed the point. You can of course increase the electric field between the plates to increase the acceleration. This will in turn mean that the potential difference across the plates has to increase (the hill has got higher). So the positive charges moving towards the positive plates have to do more work. What is gained going downhill is lost going uphill. – Farcher Feb 23 at 9:38
And as lemon mentioned in their answer, dynamically increasing and decreasing the electric field will allow you to make the ring keep accelerating, but adjusting the field in this way requires as much energy as you get out of the acceleration. – Era Feb 23 at 17:33

## protected by Qmechanic♦Feb 23 at 16:14

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