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Supposing an ideal wire, How do electrons accelerate and gain kinetic energy?

What I understand:

When a circuit is opened ,electrons are crowded at the negative term of the battery and have high electric potential energy, when we close the circuit electrons start accelerating and gaining kinetic energy through the wire.

What confuses me:

First there isn't an electric field through the wire (an ideal wire) $ {J} =\sigma {E} $ ,so I can't say that electrons gain kinetic energy and accelerate due to the electric field.

Second, if electrons lose electric potential energy (converted to kinetic energy) then there must be an electric field, because of the definition of voltage: "electromotive force, is a quantitative expression of the potential difference in charge between two points in an electrical field".

What did I miss?

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  • $\begingroup$ Well why do you think electric field can not exist in the wire with presence of battery. $\endgroup$ – Nobody recognizeable Aug 17 at 14:32
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    $\begingroup$ "Supposing an ideal wire ,How do electrons accelerate and gain kinetic energy" - why do you think there are electrons in an ideal wire? An ideal wire is an abstraction (thus the adjective ideal). $\endgroup$ – Alfred Centauri Aug 17 at 14:41
  • $\begingroup$ @Nobodyrecognizeable, Because $J = \sigma E$ ,So when conductivity is infinity the electric field is zero $\endgroup$ – Mohammad Alshareef Aug 17 at 14:50
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    $\begingroup$ @MohammadAlshareef well there's really nothing like pure conductor. Either 0/0 isn't 0. $\endgroup$ – Nobody recognizeable Aug 17 at 14:52
  • $\begingroup$ @AlfredCentauri, To simplify the problem , this how physicists do their work! why do we suppose an infinity wire or plane or anything as a whole ?! $\endgroup$ – Mohammad Alshareef Aug 17 at 14:55
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Supposing an ideal wire ,How do electrons accelerate and gain kinetic energy ?

An ideal wire is an abstraction that is used to simplify calculations in physics and electrical engineering.

The mobile charge in an ideal wire respond instantaneously to external fields so that the skin depth is zero. Clearly, this can't be the case for electrons, and so the charge carrier is 'abstracted away' in this ideal limit and we speak only of the mobile charge within the ideal wire.

In summary, if you're thinking in terms of electrons and their mass, kinetic energy, and potential energy, then it's a conceptual error to to also apply the ideal wire approximation.

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    $\begingroup$ @MohammadAlshareef, I'll make one last attempt: it seems to me that you simply don't understand the limits of the ideal wire approximation. If you have mobile electrons in your wire, it's not an ideal wire so forget about the ideal wire approximation if you ever want to understand how electrons accelerate in a wire. Does this help? $\endgroup$ – Alfred Centauri Aug 17 at 21:25
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    $\begingroup$ Since this is your last attempt,Thank you for having my back, I really appreciate you efforts, trying to help people. $\endgroup$ – Mohammad Alshareef Aug 17 at 22:35
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    $\begingroup$ If I may, a common joke in physics classes is the "spherical cow in a vacuum." We often want to use our simple equations to relate to real life situations, but nothing is so simple. As a result, we must oversimplify. There's a standing joke about a test question asking to predict the trajectory of a cow thrown from a Monty Python catapult. To avoid issues with drag or off-center rotations, the question states "assume the cow is spherical, and in a vacuum." Since then, it has become a common joke. But the issues apply here. $\endgroup$ – Cort Ammon Aug 18 at 0:07
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    $\begingroup$ If I were to ask "how hard must the spherical cow be flung in vacuum in order to exceed the speed of sound," I run into problems. The idea of a "perfect vacuum" conflicts with the idea of a speed of sound. Sound has a speed in real materials, such as air, but it has no meaningful speed in a vacuum. So to make sense of such a question, we are obliged to no longer treat the cow as a sphere in a vacuum -- we must recognize that the simple model of a sphere flying in a vacuum is no longer sufficient. $\endgroup$ – Cort Ammon Aug 18 at 0:08
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    $\begingroup$ In this case, the mere fact that you are interested in talking about electrons having kinetic energy is in conflict with the idea of an ideal wire. Its easy to come up with cases where such electrons don't make sense in an ideal wire --- Indeed, you came up with one. Thus, if you want to make sense of what is going on, you'll need to use a different model for wires, one which doesn't fall apart when you try to think about kinetics. $\endgroup$ – Cort Ammon Aug 18 at 0:09
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If you connected an ideal wire (zero resistance) across an ideal battery (no internal resistance) from Ohm's law the resulting current in the circuit would be

$$I=\frac{emf}{R}$$

Which means for zero wire resistance we would have an infinite amount of current flowing. The thing is, there will always be resistance in an electrical circuit opposing the flow of current just like there will always be some form of friction opposing the motion of masses. Both electrical and mechanical resistance (friction) result in heating.

But we do know that the resistance of wires is generally much less than the resistance of circuit components, so that we may ignore the resistance of the wires (assume them to be zero) that connect the components of the circuit. So it may be more instructive to answer your question in terms of assuming the wires that connect the circuit elements to the voltage source have zero resistance. In this case for a series circuit the current is limited only by the resistance of the circuit components. It is that resistance that prevents the charge from accelerating (thus the constant drift velocity), and where the loss of potential occurs. Since the current is the same in the wires as in the circuit elements in series with the wires, there is no acceleration of the charge in the zero resistance wires. The potential difference between the ends of the wires is zero, so there is no loss in potential.

The applicable definition of voltage here is

"The potential difference V between two points is the work per unit charge required to move the charge between the two points"

MECHANICAL ANALOGY

Consider the following mechanical analogy.

I push a box at constant velocity along the surface of a floor. The constant velocity motion of the box is analogous to the drift velocity of charge (electrical current). I, being the energy source, am analogous to the battery.

The floor surface varies such that for some stretches there is friction and the friction may vary between stretches. Let these stretches of the floor with friction be analogous to the electrical resistance of the electrical circuit components in the path of the circuit.

In between the stretches with friction there are frictionless stretches. Let these frictionless stretches be analogous to our zero resistance connecting wires.

In order for me to push the box with constant velocity on the surfaces with friction, I need to exert a force equal to the opposing kinetic friction force. The greater the kinetic friction (the greater the electrical resistance) the greater the force (strength of electric field) I need to apply and the greater the work I need to do to push the box from the beginning of the stretch to the end of the stretch. The work I do per unit mass of box pushed from a point at the beginning of the stretch to a point at the end of the stretch is analogous to the potential difference (voltage) drop across the resistor. Since there is no change in kinetic energy of the box (no acceleration of the electrons) the work I do in moving the box is dissipated as friction heating at the contact surface. This is analogous to the $I^{2}R$ heat dissipated in the electrical resistance.

I now encounter a stretch of frictionless surface (an ideal wire). No work is required for me to move the box over this surface, so I simply release it an allow it to traverse the frictionless surface at the same velocity it had at the end of the stretch with friction. That no work was needed in this stretch is analogous to there being no electric field in the stretch and no voltage drop.

When the box reaches the next stretch with friction (the next electrical resistor) I (the battery) must once again do work to keep the box (electrons) going at constant velocity (constant current).

So, to sum up, in answer to your question:

From where do electrons gain kinetic energy through a circuit?

They get it from the potential energy supplied by the battery. But they lose it when they collide or interact with the particles that the conductor is made of, such that on average, the velocity is constant (drift velocity). The energy transfer to the particles causes heating. There is always some resistance in the circuit. The parts of the circuit having much less resistance (the wires) do not dissipate the kinetic energy of the electrons, but neither do the electrons gain kinetic energy in the wires because their velocities are limited by the parts of the circuit having resistance.

Hope this helps.

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  • $\begingroup$ The claims made in the first two sentences cannot both be true. Stipulate that the claim made in the first sentence, that "there is a field in the [ideal] wire between the battery terminals", is true. Then the claim made in the second sentence, that "no work per unit charge (voltage) would be required to move the charge in the wire", cannot also be true. $\endgroup$ – Alfred Centauri Aug 17 at 17:06
  • $\begingroup$ @AlfredCentauri You are correct.The idea was to show that if both were correct the current would have to be infinite, which cannot be true. But I can see it may not come across that way. So I have revised it. Thanks. $\endgroup$ – Bob D Aug 17 at 17:20
  • $\begingroup$ @BobD ,Does $J= \sigma E$ calculate the electric field of the battery ? If yes then it should be zero because sigma is infinity (ideal wire) but you said : "If you connected an ideal wire (zero resistance) across an ideal battery (no internal resistance) there is a field in the wire between the battery terminals (due to the battery)" , how is this ? what electric field does this law calculate ? $\endgroup$ – Mohammad Alshareef Aug 17 at 18:01
  • $\begingroup$ BobD, I honestly don't get why the section OP has quoted above is even in your answer. As far as I can tell, OP is trying to understand how, when the circuit is formed, electrons can get started moving inside an ideal wire when, by all accounts, there is no electric field inside an ideal wire to accelerate the electrons. I've tried to explain, in my answer, that the abstract mobile change within an ideal wire responds instantaneously to external fields but, obviously, electrons can't do that. $\endgroup$ – Alfred Centauri Aug 17 at 19:03
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    $\begingroup$ BobD, indeed, it is questions like this and others here that have motivated me to refine my own concepts of many things including the concept of an ideal wire. $\endgroup$ – Alfred Centauri Aug 18 at 0:24

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