Cutting a polarized conductor in half

  1. 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?
     
  2. jcsd
  3. mfb

    Staff: Mentor

    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.
     
  4. tiny-tim

    tiny-tim 26,054
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    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 :wink:)

    (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? :confused:)

    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? :redface:
     
  5. mfb

    Staff: Mentor

    If you change the distance between two objects with fixed total charge, the voltage decreases with the same rate as the capacitance increases.
     
  6. tiny-tim

    tiny-tim 26,054
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    but the charge across the gap isn't fixed …

    the initial charge of ±10e on either side of the gap can change :redface:
     
  7. sophiecentaur

    sophiecentaur 13,389
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    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)
     
  8. tiny-tim

    tiny-tim 26,054
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    i don't get it :redface:

    surely the capacitance between two ends of a conductor is zero? :confused:
     
  9. sophiecentaur

    sophiecentaur 13,389
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    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?
     
  10. tiny-tim

    tiny-tim 26,054
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    i don't see the relevance of a battery :confused:

    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)?​
     
  11. sophiecentaur

    sophiecentaur 13,389
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    No it isn't. Capacitance is the ratio of the charge displaced to the applied voltage. It is the emf that has the "ability to separate charge". The Capacity tells you just how much charge the emf can displace. The charges on the wires are only unbalanced because there is an emf causing that unbalance. Movement through the Magnetic field is producing a set of series emfs along the length of the wire. I was just replacing them with a single battery so we could stop moving and look at it in detail. If you took wires from the ends and connected them across the plates of an external capacitor (with a much higher capacitance, of course) then the charge displacement would be much more. But, without that added component, there is still a displacement of charge and that implies a (very small) finite value of Capacitance.

    With circuit problems, we can usually bend and twist the actual conductors without affecting the general principles and we can replace components with other 'equivalent components' (like when we calculate parallel and series Rs and Cs) . This is what I'm attempting to do here.
    It is true to say that, if you take two equally charged capacitors and join the + end of one to the - end of the other, then the charge will be the same but the voltage across the 'outside' plates will be double. The problem with the just two wires idea is that they are not lumped components and each bit of each wire has a significant capacity between itself and every other bit. (This is unlike the example above, with two real capacitors where these extra tiny capacitances of the wires are swamped by the 01.F between the plates) So working out the actual capacitances is hard - too hard for me - but would yield the right sort of answer, I'm sure.
     
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