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Why don't electrons accelerate when a voltage is applied between two points in in a circuit? All the textbooks I've referred conveyed the meaning that when an electron traveled from negative potential to positive potential, the velocity of the electron is a constant.

Please explain.

Green Noob
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  • You are asking why is there no force on an electron from a voltage difference, i.e. from the electrostatic potential between the two terminals of your circuit. Any resulting force is proportional to the gradient of the potential. – Chris Gerig Dec 27 '11 at 04:22

5 Answers5

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Yes, the electron is accelerated by the external electric field $E$, but at the same time it is "decelerated" with collisions with obstacles. These collisions are modelled as a "friction" force proportional to the electron velocity, something like this: $$m_e\frac{dv}{dt} = eE-k\cdot v$$ This equation has a quasi-stationary solution when the dragging force cannot exceed the resistance force: $$eE=k\cdot v$$ This gives a constant (average or drift) velocity. This picture is literally applicable to the gas discharge (current in a gas) where the electrons are particles accelerated between collisions with atoms.

  • Let me ask a question: what happens when voltage (and consequently current) increases? Does the velocity increase? Or the number of free electron increases? Or it depends? – F. Jatpil Nov 16 '18 at 18:08
  • @F.Jatpil: According to the second relationship, $v\propto E$, so yes, the average velicty increases. – Vladimir Kalitvianski Nov 16 '18 at 18:58
  • my answer here completes this good https://physics.stackexchange.com/search?q=user%3A1492Why+doesn%27t+an+electron+accelerate+in+a+circuit%3Fexplanation – anna v Nov 17 '18 at 07:50
  • I can see that scattering ALWAYS produces a 'friction'/back-emf force which the current must overcome (expending energy). However I can't see the the positive lattice always absorbs this energy. e.g. if you make the positive nuclei infinitely massive it seems like you get the same 'friction' force, but there can be no energy transferred to the nuclei since they don't move... So a heavy lattice has lower resistance?! (not true) What's wrong about this picture? – Matt Aug 28 '21 at 17:10
  • Is the point that the electrons are actually accumulating a lot of randomized 'sideways' motion from the acceleration, which has to be dumped somewhere otherwise it'll grow to infinity? – Matt Aug 28 '21 at 17:12
  • And I guess that (for intinitely massive positive lattice) the scattering geometry is independent of velocity, so you are losing a constant fraction of your v^2 into sideways electrion motion, which is why P=I^2R. – Matt Aug 28 '21 at 17:26
  • So the 'force field' producing the 'back-emf' of a resistor is actually from whatever force damps the sidewarys motion of electrons -- i.e. the back-emf acts perpendicular to the battery EMF! But since the electric field lines in a resistor are wiggling about a bit due to superposition of the scattering-object field, a perpendicular field can legitimately act against a driving EMF. – Matt Aug 28 '21 at 17:31
  • Free electrons collide also with bound electrons. There is a lot of obstacles for a free electrons, as a matter of fact. – Vladimir Kalitvianski Aug 29 '21 at 06:50
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Electrons are accelerated by the constant applied electric field that comes from the external potential difference between two points, but are decelerated by the intense internal electric fields from the material atoms that makes up the circuit. This effect is modeled as resistance.

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And one can add that they are accelerated if the circuit element and the voltage difference is the one applied on a vacuum tube, the simplest particle accelerator. In the vacuum there is no resistance and statistical transfer of energy to other electrons.

anna v
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  • I think this should be a comment on another answer. Or you can complete your answer instead. – FGSUZ Nov 16 '18 at 17:17
  • @FGSUZ if you note the date, this happened seven years ago,when I was new on the site. – anna v Nov 16 '18 at 18:16
  • Opps. That's right haha. I saw this on "Q's with new activity" and thought it was new. It remains open however, so I guess it's never too late haha, but just up to you. – FGSUZ Nov 16 '18 at 21:41
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You may consider the relation: $$i=nAev_d$$ $$=>v_d=\frac{i}{nAe}$$ $A$ : Cross -Sectional Area

$n$ : Concentarion of electrons

$v_d$ : Drift velocity

$i$ : Current

If a constant current flows through a conductor of varying cross section the drift velocity will change

In fact we have the relation:$$j=\sigma E$$ If the cross section changes[current remaining constant ] the current density,$j$ will change. Consequently E will change if the conductivity $\sigma$ is constant[for a homogeneous material with constant values for n and $\sigma$].

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At the classical level the explanation provided in the previous answers is known as the Drude model. There is additional info on Wikipedia: http://en.wikipedia.org/wiki/Drude_model

Whelp
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