# Cutting a polarized conductor in half

by Artel
Tags: conductor, cutting, polarized
 P: 6 If you have a bar conductor that's moving perpendicular to a magnetic field, the magnetic force would bring all the electrons to one side of the conductor, leaving one end positive and the other end negative. If you cut the conductor in half while it is moving, will one end be positively charged and the other end be negatively charged?
 Mentor P: 9,636 The magnetic force does not bring all electrons to one side - actually, the imbalance will be extremely small for realistic setups. If you cut the conductor, the parts will be charged afterwards, right.
 PF Patron HW Helper Sci Advisor Thanks P: 25,517 let's look at it the other way round … suppose you start with two bars aligned in a field, so that each bar has 10 electrons at the left end and 10 "holes" at the right end now gradually bring the two bars together … as they get closer, more electrons from the right bar will be attracted to the left end of the right bar (that's what would happen if the left bar was an insulator with a positive charge at the right end, in a zero field ) (to put it another way: the voltage that should be across the two bars is entirely across the gap between them: as the distance decreases, the capacitance increases, so the charge must increase! … hmm, why doesn't it become infinite? ) just before they touch, there will be 20 electrons at the left end of the right bar (and 20 "holes" at the right end of the right bar), and 20 "holes" at the right end of the left bar (and 20 electrons at the left end of the left bar) when they touch, the 20 electrons and "holes" in the middle will join up, and disappear, leaving only 20 electrons at the left end and 20 "holes" at the right end (of the complete bar)! have i got that right?
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## Cutting a polarized conductor in half

 Quote by tiny-tim (to put it another way: the voltage that should be across the two bars is entirely across the gap between them: as the distance decreases, the capacitance increases, so the charge must increase! … hmm, why doesn't it become infinite? )
If you change the distance between two objects with fixed total charge, the voltage decreases with the same rate as the capacitance increases.
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 Quote by mfb If you change the distance between two objects with fixed total charge, the voltage decreases with the same rate as the capacitance increases.
but the charge across the gap isn't fixed …

the initial charge of ±10e on either side of the gap can change
 PF Patron Sci Advisor P: 10,118 I tried looking at the problem this way and I can't solve it but it may be a way in: The total induced emf (V) in the two moving wires will be the same, whether joined at the centre or not, I think, because d∅/dt will be the same. They will 'share' the induced V. The charge polarisation will be V/2C. where C is the capacitance between the two ends of a single wire. When the wires are joined together, the d∅/dt for the double length wire will be twice but the capacity will not be simply half of the capacity of one short wire (will it?). If it were, then the total charge imbalance would be identical to the charge on each wire. I have ignored the process of actually joining the wires together as there seems to be an unexplained change in energy (as with the classic 'paradox' of joining capacitors in parallel)
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 Quote by sophiecentaur … where C is the capacitance between the two ends of a single wire.
i don't get it
surely the capacitance between two ends of a conductor is zero?
 PF Patron Sci Advisor P: 10,118 Well. If you insert a battery half way along it, there will be a certain displacement (Q) of charge so this represents a Capacitance, given by C=Q/V. Bearing in mind that 'everything' has capacitance and inductance, even if it doesn't actually have plates andcoils in it, then I feel it's all right to use Capacitance in this instance. I'm afraid that I don't know how to calculate it but it must relate in some way to the length of the wire and you can bet that it won't be simply proportional to the length. This is 'reasonable', isn't it?
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 Quote by sophiecentaur Well. If you insert a battery half way along it, there will be a certain displacement (Q) of charge so this represents a Capacitance, given by C=Q/V. Bearing in mind that 'everything' has capacitance and inductance, even if it doesn't actually have plates andcoils in it, then I feel it's all right to use Capacitance in this instance. I'm afraid that I don't know how to calculate it but it must relate in some way to the length of the wire and you can bet that it won't be simply proportional to the length. This is 'reasonable', isn't it?
i don't see the relevance of a battery

capacitance is the ability to separate charge

two copper plates separated by vacuum will have a certain capacitance, C … apply a voltage, V, across them (eg by connecting a battery "the long way round"), then positive charge CV will build up on one plate, and negative charge CV on the other

fill the gap with a dielectric, that will increase C, and the charge will increase

but fill the gap with copper, and no charge will build up … the copper plates are now two ends of a conductor, and they can't separate charge
surely capacitance in this case is only relevant when there is a gap in the conductor (the wire)?
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