Why doesn't an electron lose energy between components?

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shrey012
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I understand that in an electrical circuit, a positive test charge will gain electrical potential energy when moving from the negative to positive terminal of a battery (as it's moving against the electric field); it will then naturally flow around the circuit back to the negative terminal as it "falls back" from a higher electrical potential to a lower electrical potential.

However, I cannot see why an electron (or "positive test charge" if we stick with conventional current flow) doesn't dissipate energy as it flows to the lower potential terminal between components - i.e. when the electron isn't traveling through any circuit component, but just the wire. For example, I know the electron will transfer energy to heat as it moves through a resistor, or light and heat as it moves through a filament lamp, but why would it also not be losing energy between these components?

If you raise a ball to a particular height and let it fall, it immediately starts dissipating the gravitational potential energy (into kinetic energy). You can also transfer the energy into other forms if you put things along its path (e.g. a viscous liquid to transfer some of the energy to heat) - these additional "things" could represent the circuit components. However, between the "circuit components", the ball will still be dissipating energy.

Lastly, once an electron has transferred all its energy after it passes through the last component, how is it still able to move to the lower electrical potential terminal of the battery?

Thank you very much.
 

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  • #2
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Energy is not a sort of currency that is owned by individual charge carriers and used to pay tolls as it travels. The energy is stored in and transferred by the fields.

If energy were carried by the charge carriers then it would take many minutes between when you close a switch and when a light bulb would light. The charge carriers move very slow, the energy moves very fast, as fast as the fields change, this shows that the energy is carried by the fields not the charge carriers.
 
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  • #3
ZapperZ
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However, I cannot see why an electron (or "positive test charge" if we stick with conventional current flow) doesn't dissipate energy as it flows to the lower potential terminal between components - i.e. when the electron isn't traveling through any circuit component, but just the wire. For example, I know the electron will transfer energy to heat as it moves through a resistor, or light and heat as it moves through a filament lamp, but why would it also not be losing energy between these components?

If you raise a ball to a particular height and let it fall, it immediately starts dissipating the gravitational potential energy (into kinetic energy). You can also transfer the energy into other forms if you put things along its path (e.g. a viscous liquid to transfer some of the energy to heat) - these additional "things" could represent the circuit components. However, between the "circuit components", the ball will still be dissipating energy.

Lastly, once an electron has transferred all its energy after it passes through the last component, how is it still able to move to the lower electrical potential terminal of the battery?

Thank you very much.

The problem with your analogy is that we don't quite have something with "negative mass". If we do, then it will "fall up" to get to a lower energy state. The closest we have in terms of that analogy is a bubble (which, if you know quantum field theory and condensed matter physics, is considered as the positive holes in a conductor/semiconductor). A bubble wants to go up, because that is where it will have a lowest potential energy. Similarly, for helium and hot-air balloon.

Zz.
 
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Drakkith
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If you raise a ball to a particular height and let it fall, it immediately starts dissipating the gravitational potential energy (into kinetic energy).

That isn't what "dissipate" means. Its definition is: "cause (energy) to be lost, typically by converting it to heat". The potential energy of the ball is converted to kinetic energy during the fall, however. I don't say all this to be nit-picky, but because energy in a circuit is lost in a non-conservative manner, unlike the conversion of gravitational potential energy into kinetic energy (a conservative process because gravity is a conservative force).

Lastly, once an electron has transferred all its energy after it passes through the last component, how is it still able to move to the lower electrical potential terminal of the battery?

There is still a potential difference between the last component and the battery or other voltage source (regardless of whether or not you can think of this in terms of single electrons, which you really can't).
 
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CWatters
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However, I cannot see why an electron (or "positive test charge" if we stick with conventional current flow) doesn't dissipate energy as it flows to the lower potential terminal between components - i.e. when the electron isn't traveling through any circuit component, but just the wire. For example, I know the electron will transfer energy to heat as it moves through a resistor, or light and heat as it moves through a filament lamp, but why would it also not be losing energy between these components?

If you treat wires as ideal conductors (zero resistance) then both ends of a wire between components is at the same potential. In practice real world wires behave like low value resistors, they do have a small resistance and so there is a small voltage drop.
 
  • #6
vidit jain
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Consider the analogy with a falling ball.
When a ball falls in the atmosphere, it dissipates heat in the form of heat,in a non conservative manner.
But when it falls thru vacuum, no heat is released.
An ideal wire is analogous to a vacuum pocket; when the electron 'falls' through the wire, it doesn't lose any energy.
Other components of the circuit which provide resistance maybe compared to the atmosphere;both cause the respective bodies to lose energy.

Note: the wires used in everyday circuits have a small resistance . But there exist materials which pose 0 electrical resistance to flowing current,namely superconductors. The whole of this thread isn't enough to encompass the awesomeness and the wonders of superconductors.
 
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Drakkith
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Consider the analogy with a falling ball.
When a ball falls in the atmosphere, it dissipates heat in the form of heat,in a non conservative manner.
But when it falls thru vacuum, no heat is released.

That's a interesting analogy. A steady current through a non-ideal wire can be thought of as balls falling at terminal velocity then.
 
  • #8
David Lewis
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...energy in a circuit is lost in a non-conservative manner, unlike the conversion of gravitational potential energy into kinetic energy (a conservative process because gravity is a conservative force).

That is true if you're talking about a resistor or light bulb. Electrical energy is conserved when converted to potential energy stored in a magnetic or electric field.
 

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