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The electric field is given by: $$\vec{E} = \frac{1}{q}\vec{F}$$ Electric charges create an electric field which, in turn, creates a force that accelerates charges.

The Ohm's law, however, tells us that:$$\vec{j}=\rho\vec{v}=\sigma \vec{E}=\frac{\sigma}{q}\vec{F}$$ with $\vec{v}$ being the velocity of one charged particle.

My question is this: if electric fields cause charges to accelerate, how can electric currents be defined as proportional to a constant velocity? Is the $\vec{v}$ in the equation above constant at all?

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  • $\begingroup$ We assume that the electrons move with constant speed called as drift speed. $\endgroup$
    – Shreyansh Pathak
    Sep 25, 2019 at 13:54
  • $\begingroup$ So it's just an approximation? What is the actual connection between the electric field and the electric current? $\endgroup$
    – Hrach
    Sep 25, 2019 at 13:56
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    $\begingroup$ Well, actual scenario can be explained using quantum mechanics. $\endgroup$
    – Shreyansh Pathak
    Sep 25, 2019 at 13:59

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The Ohm's law was concluded from what was found from experiments. It is true that "electric field cause charges to accelerate", but under conditions the Ohm's law apply (e.g., in conductors), there are a whole lot of other forces that affect the movement of electrons. The collisions of a moving electron to non-moving atoms serve as friction force which slows down the moving electron.

Consider this analogy. When you apply force to a balloon, it moves. But it can not keep accelerating to very high speed because friction increases with its speed. It has certain final speed corresponding to certain force. With small forces the formula may be $$\vec{v}=k\vec{F}.$$ You may ask, How can speed $\vec{v}$ be constant with constant $\vec{F}$ for the balloon? well, the answer is speed-dependent friction. You do not need to resort to quantum mechanics (as suggested by a comment) to understand it.

So both of your two equations are correct, but they are correct under different conditions.

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  • $\begingroup$ But even if we assume perfect vacuum where Ohm's law doesn't work anymore, there is a connection between electric field and current, correct? In which case, if there is an electric field present the electric current is not constant anymore and the charges are being accelerated. Also how can one derive the connection between field and current? $\endgroup$
    – Hrach
    Sep 25, 2019 at 14:39
  • $\begingroup$ @ConnyDago In a vacuum tube there is vacuum and Ohm's law does not work. There the current is determined by some other factors, including field strength, number of electrons available (heated filament is used to emit electrons which has certain capability), distance between anode and cathode, structure details, etc. Search for "vacuum diode" online, you will find explanations. $\endgroup$
    – verdelite
    Sep 25, 2019 at 14:47
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An electric charge placed in an electric field will accelerate if there is nothing to slow the charge down. This would be the case if the electric charge were placed in vacuum where an electric field existed.

Ohms law pertains to conductors and all real conductors have resistance. At the same time the field accelerates the charges giving them kinetic energy, the charges in turn collide with the atoms, molecules and the like of the material giving up kinetic energy. That kinetic energy winds up as heat in conductor. This occurs such that, on average, the kinetic energy lost in collisions equals the kinetic energy gained by the field, so that on average the change in kinetic energy is zero and average velocity (drift velocity) of the charges is constant.

Hope this helps.

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A charged particle when placed in a uniform electric field will undergo acceleration as long as there are not obstacles in its path.

What you are referring to as the velocity of electric current is actually drift velocity. Drift velocity is the result of the electron repeatedly accelerating, colliding with another electron, coming to a stop and accelerating again. This when averaged this gives the velocity with which the electron could be said to be moving at. This velocity can be derived by considering the electric field, and relaxation time of the electron. The relation comes out to be $$v_d = {eEt\over m}$$

where e = charge of electron, E = electric field intensity, t = relaxation time, m = mass of electron

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