# Why doesn't an electron accelerate in a circuit?

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.

• 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 4:22

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 physics.stackexchange.com/… – anna v Nov 17 '18 at 7:50

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.

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.

• 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

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 $\sigma$].

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