 # Refuting the laws of physics (II)

Man creates laws of physics to describe reality. But then he gets so much attached to them that he often forgets that theory ought to obey the facts and not viceversa. Continuing my struggle against this "scientific" idolatry initiated with my article on hydrogen atom in normal state, I would like to consider with your help, dear reader, the legitimacy of the claim that

## Electrostatic field does not penetrate into a conductor.

Well, you may ask now, what about Ohm's law? In the presence of tension, electric field is unavoidable. Of course, it is not my intention to catch you on such obvious things. I am thinking of an electrostatic field, i.e. a field which exists in the state of static equilibrium when there is no current flowing. No current indeed? For you to judge.

The Proof is simple. There can be no electric field in the conductor, for it there were one, the motion of its free charges caused by a force acting on them would make the field shrink (see Fig. 1). Fig.1. Conductor in an electrostatic field.
a) in an activated field
b) in the state of equilibrium

The motion will persist as long as there is any field in the conductor. When the field disappears completely, a static situation is attained - the state of equilibrium.

## And yet it does penetrate

To see this, let's take a closer look at the conductor. The situation presented in Fig. 1, with free charges of either sign, is quite rare (e.g. the plasma formed by an ionized gas). Most frequently, charges are all of the same sign (in metals - negative electrons), while compensating charges of opposite sign (in metals - positive ions) remain fixed in the conductor, forming a periodic space lattice (crystals) or a chaotic system (amorphic bodies). If an electric field is applied so as to push the free charges into the conductor, the surface layer of the charge which compensates the field within it is due to a fixed charge, uncovered by the retractation of free carriers. The surface density of the charge is determined by the electric field surrounding the conductor and can be computed by Gauss' theorem. To attain that surface density, free charges must be moved back to a depth sufficient to uncover the required number of fixed ions of opposite sign. A charged layer of thickness is thus formed (where \Sigma is the surface density and \rho_0 the volume density of the fixed charge). Within this layer the field fades away from its maximal value to zero (Fig.2). Fig.2. Fading of the electric field in a conductor with negative current carriers (field directed outwards).
a)The field with movable ( - ) and fixed ( + ) particles.
b)Field intensity dependent on the distance from conductor surface.
c)Electric charge density dependent on situation.

We have thus proved that the field penetrates into the conductor. Let's estimate the thickness of the layer under the assumption that the electric field intensity at the metal surface is (typical for electrostatic experiments) and copper is the conductor - volume density of the charge is then given by Then the thickness will be equal to Now you may exclaim: This is absurd! Why should I ever bother about a layer devoid of carriers, the thickness of which is moreover neglectably small with respect to atom dimensions? This is all true, but take a look at semiconductors. For them, the concentration of free carriers being less by several orders of magnitude than in metals, the penetration depth of the field may be significant. What is more, this fact makes possible volume density measurement of the fixed charge and it even led to the emergence of a new experimental technique known as capacitive spectroscopy. This technique is used to investigate the evolution of admixtures in semiconductors in charge state. Why capacitive? Because the layer devoid of free carriers forms a condenser of a sort. By measuring its capacity we can determine the thickness of the layer and hence also the density of the charge. How? Let's leave it for some other time.

All this may be true, you may object, but this is not a conductor. A layer without carriers does not conduct electricity, so no wonder the field gets inside. We have refuted nothing so far! Well, if this is not sufficient evidence for you, I am not giving up yet. We shall now see that

## The field penetrates into a conductor too

To this aim let's have a look at the other end of the conductor considered.

No problems to be encountered there, you may say, free charge can reach the surface and form a layer compensating the electric field. Using this argument, we forget a frequent phenomenon which smooths all irregularities: diffusion.

It is well known that when the concentration of movable particles is irregular, they are displaced so as to diminish the degree of irregularity. This process is described by the diffusion equation, which in the one-dimensional case has the following form: where j_d is the diffusion current, i.e. the number of particles flowing through a unit area perpendicular to the mean direction of their motion in unit time, n is the concentration of particles (number of particles per unit volume) and D is the diffusion constant.

Diffusion leads to the erosion of the immense (theoretically infinite) volume density of the charge concentrated on the surface, needed to prevent the field from penetrating into the conductor. In turn, the erosion of the charge on the surface allows the field to enter into the conductor to some depth. The field, by pulling the carriers towards the surface, stops the diffusion process, thus establishing equilibrium (Fig.3). Fig. 3. Fading of the electric field in a conductor when the field is directed inwards (notation as in Fig.2)

If the qualitative argument does not satisfy you, you may ask

## How deep does the field penetrate into the conductor?

To find the equation showing how the field depends on its position x in the semiconductor, we must require that the electric current emanating from the field be compensated by the diffusion current. The sources of these two currents are related to charge distribution, so we may hope the above condition will make possible the computation of the distribution.

The value of the diffusion current must be equal to minus the density of the electric current caused by the field, divided by the charge of a single carrier. But Volume density of the free charge being related to carrier concentration n by the equation $\rho = qn$, we get Since we arrive at On the other hand, where $\sigma$ is the specific conductivity of the material.

Substituting (2) and (3) into (1) we now have i.e. Observe that the dimension of the constant is the inverted square of some distance. Call this distance x_0. Then, obviously, Thus the electric field must fade away according to the equation It can be easily verified that this equation is satisfied by an exponentially fading field: where x_0 is the distance over which the field decreases e times. Thus, it characterizes the penetration depth of the field into the conductor.

Substituting values known for copper: and assuming that epsilon is equal to 1, we get Now you will accuse me of pulling your leg by suggesting that field penetration with depth comparable to the dimension of an atom can be treated seriously. Well, my answer is that you are absolutely right. However, in semiconductors, for which specific conductivity is by far smaller, whereas diffusion is by far greater than in metals, the distance x_0 may become significant. Moreover, please note that our entire argument was based on laws of classical physics. It seems that we have forgotten that in the microworld these laws should be replaced by those of quantum mechanics. The latter describe the reluctance of particles to gather in a small area. Nevertheless, you may believe me when I say that the effects considered above would remain unaltered had the correct laws been taken into account.

It's time to ask again: what has been refuted above? A law of physics or a commonplace belief? It's for you to judge, dear reader.

Jan A. GAJ
Department of Physics, Warsaw University 