Let's assume we have a circuit without a battery,the wire in the circuit is not an ideal wire and the free electrons will have to do work to cross them. Now let's assume that as there is no battery connected electric potential at a point is $0$. As it is not an ideal wire the electrons will have to do work and hence there will be a loss in energy let's call that $U$. Now, as $\displaystyle V_o-V_f=U×(-q)\implies V_f=Uq+Vo$. Hence the potential at the point ahead of our original point is positive and electrons will get attracted to it. But this doesn't happen in real life why?
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The work done is the potential difference times the charge. – Gilgamesh Jun 16 '20 at 06:33
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A simpler example to disprove perhaps: your reasoning seems to also suggest that a red LED should always have 1.7 volts across it, for the same reason. (therefore if I connect a blue LED to a red one, the red one should glow) – user253751 Jun 16 '20 at 19:48
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As it is not an ideal wire the electrons will have to do work and hence there will be a loss in energy let's call that U.
The electron has to be supplied energy to move , which is done by the battery, the solid itself cannot give it. Your mistake is in energy conservation violation
Because of energy and momentum conservation there will be no drift velocity for your circuit , no moving electrons. When a voltage is applied a drift velocity can be defined.
To describe correctly the behavior of electrons,atoms, molecules quantum mechanical equations and postulates have to be used. Electrons are not free to move like billiard balls. In general they are tied up with binding energy to atoms, in a quantum mechanical energy level.
In solids , where there are many atomic and molecular interactions , a generalized quantum mechanical model has been found to fit the data, called the band theory of solids.
An important parameter in the band theory is the Fermi level, the top of the available electron energy levels at low temperatures. The position of the Fermi level with the relation to the conduction band is a crucial factor in determining electrical properties.
In the circuit you posit the conductors are in a lattice and part of the electrons are shared with the whole lattice. In quantum mechanics it means that an electron in the conduction band has a probability of being in any part of the conductor, but no motion is involved. If there is no battery the specific location will be anywhere
anna v
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So does it mean that if we do not have the energy present the point Vf does not have higher potential.It will have only higher potential when the energy will be supplied? – BlackSusanoo Jun 16 '20 at 07:48
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Adding to Anna v answer. I think that as there is no battery in the circuit there would not be any electric field, electrostatic potential energy drops only when there is an electric field. So potential at the points would be zero.
BlackSusanoo
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