(constants and dimensions)

The production and measurements of electricity started with the
invention of the battery by Alessandro Volta. Batteries are
usually made of a combination of successive substances (different
metals separated by an electrolyte solution), so that the movement
of each electron through the battery is "pushed" by a chemical
reaction (each atom of zinc dissolved pushing 2 electrons). Each
specific combination of chemicals (metals) determines a difference
of electric potential between two electrodes, independently of the
size of things. This is how was initially defined the unit of
electric potential, the Volt: it came from the difference of
energy between different chemical combinations of atoms that we
had under hand. But we now commonly have batteries giving about
1.5 V.

We can refer to a more fundamental process that detemines the
order of magnitude of such potentials, though it cannot be
directly used in a battery : from an isolated hydrogen atom in its
lowest energy level, the energy needed
to take away its electron is equal to 13.6 eV (we shall
precisely describe this process below) where 1 eV = 1 electronvolt
= the energy needed to carry 1 electron across 1 volt. So,
batteries can only provide for each electron a small fraction
(about 10%) of the "fundamental" quantity of chemical energy per
electron.

Then came the definition of the unit of electric current
(amperes), as derived from the units of electric potential, time,
and energy: the electric power P in watts (= joule per second,
where 1 joule is twice the kinetic energy of 1kg moving at the
speed of 1 m/s) exchanged by an electric current *I* passing
through an electric potential (voltage) difference of *V* is
P=IV.

So, 1 ampere of current going through 1 volt of potential,
dissipates 1 joule per second. Unlike the volt, the choice of the
ampere as a unit of current does follow any fundamental (naturally
occuring quantity), but its order of magnitude depends on our
conventional units of mass, space and time. We will try to
describe later how big it is.

What we can already say, is that a voltage V between two ends of
a resistor with resistance R creates an electric current I=V/R
through the resistor. The power dissipated through the resistor is
thus P=V^{2}/R.

But the value of the resistance depends not only on the type of
material but also on the dimension. Denoting *ρ* the
electrical resistivity of the material, x its length and A its
cross-sectional area, its resistance is R= ρ x/A so that P=V^{2}A/
*ρ* x.

This is because the electrons are attracted to flow at a speed that is proportional to the electric field in the object. The unit of electric field is V/m (volt per meter). This is the equivalent of a slope, while the role of the electric potential is played by the altitude. A difference of potential is represented by a difference of altitude, so that the same difference of potential can either take the form of a smooth slope on a large distance, or a steep slope on a short distance. A steep slope pushes things downwards more strongly than a smooth one. So the intensity of the electric field is E=V/x while the current is attracted by it in parallel in all the cross-sectional area : I = AE/ρ=AV/xρ, so that the ratio defining the resistance R=V/I is given by ρ x/A.

Since A is a surface while x is a length, a resistivity is homogeneous to a resistance multiplied by a length, so that for a given voltage and a given type of material, the power of a cubic resistor is proportional to its size. But, there is a big range of variation between the resistivities of different materials. Is it because the ones are very big, or that the others are very small ???

Electricity does not just flow around in closed circuits. It is a
flow of charges subject to the electrostatic interaction.

If we only had to deal with static charges related with mechanical
interactions, we could choose a unit of charge so that the
mechanical strength of this interaction seems "normal" (neither
big nor small) in our everyday experience.

The problem is that we must also relate this with our units of
electricity.

So we define the unit of charge, the Coulomb, as the charge
carried by a current of 1 ampere each second. As 1 ampere
dissipates 1 joule per second across 1 volt of potential, it means
that 1 coulomb is the charge that takes or disspates 1 joule when
crossing a difference of potential of 1 volt (no matter how much
time it takes).

The relation between quantities is :

Energy = Charge * Potential

Charge = Energy / Potential

1 Coulomb = 1 joule/volt

Charge = Energy / Potential

1 Coulomb = 1 joule/volt

But, how big is 1 coulomb ?

If we take an elementary charge e=1.6× 10^{−19} C
and multiply it with the Avogadro number 6.022 × 10^{23},
we see that a charge of 96000 coulombs, is (minus) the charge of 1
mol of electrons (thus 1 gram of protons).

Thus an electic current of 1 ampere takes 96000 seconds (about 27
hours) to carry 1 mol of electrons, thus to make use of 1 mol of
substance in a battery (a battery with this quantity of substance
may have a weight of 100 grams or the like). Fortunately it lets
us a significant time of use of such a battery before it is
finished.

Seen in this way, 1 coulomb might look like a small charge.
But... wait a minute.

This process is made slow by our definition of the ampere: it
must give "only" 1 joule per second across the potential of 1
volt. But 1 joule is a relatively small energy as it is twice the
kinetic energy of 1 kg going at "only" 1 m/s which is a very slow
speed for fundamental physics. But we satisfy ourselves with such
small energies at a time, while a whole battery contains a lot
more.

Now let us measure the electrostatic interaction with respect to
electrical quantities. Assuming this interaction to take place in
the void, it is given by the electric permittivity of the
void ε 0
= 8.8542×10^{−12}
C.V^{−1} .m^{−1} = 8.854×10^{−12}
C^{2}.J^{−1}.m^{−1}.

This way, the interaction between 2 charges q and q' can be
expressed by the potential energy qq' /(4πε 0r).
Thus the potential energy of electrostatic interaction beween 2
charges of 1 coulomb 1 meter apart, is... 8.99 × 10^{9}J
, to be compared to 1
barrel of oil = 6.118 × 10^{9}J.

We can also write ε 0 = 8.854×10^{−12} F·m^{−1}
where F is the Farad, that is the unit of capacitance, 1F= 1C/V

A capacitor is an electric device that does not let the electric
charge go through but accumulates it on a surface while an equal
amount of charge goes out on the other side, from a close
(parallel) surface. Thus it keeps opposite amounts of charge to
face each other. It is shaped like a resistor that would be
designed to have the smallest possible resistance but made from a
material with high electrical resistivity. Thus if we don't want
the current to go through in a capacitor, we need to make it with
a material with even higher resistivity, or possibly the void.
Taking the void to separate the surfaces, the capacitance of a
device made of 2 conducting surfaces with area A separated by a
distance x is ε_{0} A/x. This is the ratio of the quantity
of charge stored in one surface (with the opposite charge on the
opposite surface) by the difference of potential between them.

While an electric current flowed through a resistor, guided by
the electric field, all what we have in the capacitor is the "flow
of the electric field" without any electric current following it
because the material (or the void) does not allow it.

The analogy can be taken seriously as the Maxwell's equations
directly express an abstract definition of a "current"

j+ ε_{0} ∂E/∂t

As a capacitance is a number of coulombs per volt, and a coulomb is a joule per volt, we can directly write the energy containted in a capacitor as : energy = capacitance×potential

Thus we need to make A/x as big as possible to allow for a more significant quantity of charge, and thus of energy, to enter a capacitor when the potential is fixed.

Let us further compute the energy of a capacitor :

Energy = (ε_{0} A/2x)×potential^{2} = ε_{0}
volume× (potential/x)^{2}/2.

Thus the energy density of the capacitor is ε_{0}(potential/x)^{2}/2,
where (potential/x) is the intensity of the electric field.

But ε_{0} is numerically very small. Thus in the way we
introduced (measured) it, the electric field is very weak as
concerns its energy density. For it to store more energy, we need
a much bigger electric field. Either with a constant voltage by
taking a very small x (to microscopic size) or a high voltage.

However this has a physical
limit. The dielectric strenght of the air (the value of the
electric field at wich a disruptive charge happens) is of 3.6
megavolts per meter. The energy density of this field is
8.85e-12*3.6e6^2/2= 57 J /m^{3}. The same density of
energy can be found in the form of kinetic energy of flowing water
at the speed of 35 cm/s. As the density of paper
is about 750 kg/m3 (= 3/4 that of water), this energy density is
the one that can lift a slice of paper with width 57/(9.81*750)
meters = 7mm

I don't even dare a comparison with the energy density of oil.

Finally, how big is 1 coulomb ?

We can also write ε 0 =
8.854×10^{−12} C m^{-2}
(V/m)^{-1}.
This means that the dielectric strength of the air is that of the
field generated by a electric charge of 3.1×10^{−5}
coulombs per square meter. In other words, if there is air around,
1 coulomb must be spread across at least the area of a 190 meters
wide square (and not be contained in a sphere with this area) to
not flee as a disruptive charge. "An average bolt of negative
lightning carries an electric current of 30,000 amperes (30 kA),
and transfers 15 coulombs of electric charge and 500 megajoules of
energy. Large bolts of lightning can carry up to 120 kA and 350
coulomb" (wikipedia).

Now let us finally assess how big are resistivities of different materials. Given the resistivity ρ of a material, the quantity ε0ρ in the international units system, is a number of (C/Vm)(Vm/A)=C/A= time (seconds). This quantity is the range of magnitude of the time it takes for a solid made of this material to approach its stable distribution of charges (say, become twice closer to it), if it is compact enough (like a sphere or a cube, but not something stretched).

So, let us now copy part of the table of resistivities from
Wikipedia and express these values in quantities of time:

Silver | 1.4 × 10^{-19} s |

Lead | 2×10^{−18} s |

Carbon (diamond)^{} |
8.85 s |

Sea water^{} |
1.7×10^{−12} s |

Drinking water | 2×10^{-10} to 2×10^{-8} s |

Silicon^{} |
5.66×10^{-9} s |

Deionized water^{} |
1.6×10^{-6} s |

Glass | 1 to 10000s |

Hard rubber | 88 s |

Sulfur | 8854 s |

Air | 32 to 80 hours |

Paraffin | 250 hours |

Fused quartz | 76 days |

PET | 280 years |

Teflon | 10×10^{22} to 10×10^{24} s |

But 1Ω×ε 0 = (8.854×10

Its inverse is thus 1.13× 10

Let down electrostatics. The main effect that can happen in
ordinary electrical circuits that makes their behavior differ from
basic instantaneous equations of electricity, is the magnetic
effect.

The amplitude of magnetic effects is characterized by the constant
μ_{0},= 1/(ε 0c^{2})
= 4π×10^{−7}
V·s/(A·m) (where 4π×10^{−7}=1.256×10^{−6})
commonly called the vacuum permeability, permeability of free
space, or magnetic constant.

It is a relativistic effect of electrostatics.

Relativistic effects happen when some speed v approaches the
speed of light c, and the range of magnitude of these effects is
given by (v/c)^{2}.

Magnetism is the relativistic effect produced by the movement of
electrons inside electric wires. Not the movement of these
electrons around the nucleus, nor the thermal agitation (that will
be computed later).

What approaches the speed of light here, is the part of the
movement of electrons inside electric wires that make the electric
current.

Let us refer to the definition of the
Ampere : "the constant current that will produce an
attractive force of 2 × 10^{–7} newton per
metre of length between two straight, parallel conductors of
infinite length and negligible circular cross section placed one
metre apart in a vacuum".

Consider a wire of copper 1mm wide transporting an electric
current of 1 ampere. Its mass density is 8.96 g·cm^{−3}.
Its standard atomic weight is
63.5, so that its density of substance is 8.96/63.5= 0.14
mol·cm^{−3}. In other words, 1 atom per volume of a
cube with size, (0.14×6.022×10^{23})^{-1/3}/100=2.28
× 10^{-10} m, or 2.8 Angstrom. In fact, in copper,
only one electron per atom participates in the current.

Each second, this current carries 1 coulomb, that is about 10^{-5}
mol of electrons. A copper wire with section area 1mm^{2}
would carry them on a distance of 10^{-5} mol /
(0.14 mm^{2} mol·cm^{−3})= 0.07 mm.

Is it slow ? Yes it is. And the magnetic force it generates, is
quite small. We can make it bigger by putting 100 such wires in
parallel (that can be the same one making a big loop), bringing 1
ampere each, for a total of 100 amperes across a section of 1cm^{2
}made of 100 wires of 1mm^{2} each. Then the speed of
electrons remains the same, but the total force between 2
such things one meter apart is multiplied by 100 ×100, thus
giving a force of 2 × 10^{–3} N/m. And if
you put 10 amperes in each wire instead of 1 (thus a speed of
electrons of 0.7 mm/s), and put these 10cm apart instead of 1 m,
you get a force of 2 N/m. Wow ! this is already significant.

Now here is the explanation of the involved relativistic effect at this speed.

With respect to the ground, each wire can be abstractly
understood as a superposition of a system of steady ions of
copper, with a system of flowing electrons. Both have the same
(opposite) density of charge d.

Wire A = Copper A + Electrons A

Wire B = Copper B + Electrons B

With respect to the ground, the repulsive force between Copper A
and Copper B is cancelled by the attractive force between Copper A
and Electrons B. Then, we have the same amount of attractive force
between Electrons A and Copper B.

But, how much is the repulsive force between Electrons A and
Electrons B ?

This is where the relativistic effect comes into play. Assume they
go at the same speed in the same way.

One meter (with respect to the ground) of Elecrons A going at
speed v, that are made of the same number of electrons as the
copper charges, see in their own perspective, Electrons B as less
dense per meter (a multiplication by √1−
v^{2}/c^{2} ),
in order for the relativistic contraction of lengths to give them
back the density d from the ground's viewpoint. Moreover, this
weaker electrostatic force, that is an exchange of momentum per
unit of time, is calculated in the Electron A's time that has a
relativistic time dilation with respect to the ground. Therefore,
the expression of this force as an exchange of momentum with
respect to the ground's time, provides another multiplication by √1− v^{2}/c^{2},
leading to a total decrease by v^{2}/c^{2} times
the initial amount.

This slight relative modification affects an abstract value of an
electrostatic interaction between systems of many coulombs of
charge. You can compute yourself how big this abstract value is,
and this way understand that even a small fraction of variation
produces a significant difference when the initial value is
cancelled by the interaction between Electrons A and Copper B.

Let us now compute the energy density of the Earth's magnetic
field.

It is B^{2}/2μ_{0} where the Earth's magnetic
field B = 25 to 65 microteslas. If we take 50 microteslas we
have (5×10^{−5}
)^{2}/(2×4π×10^{−7}
)=10^{−3} J/m^{3}.

There are much stronger magnets than this; unlike electric fields,
there is no theoretical limits to the intensity of a magnetic
field. Note that the intensity of the magnetic field at the
surface of a magnet is independent of its size. (but depends on
its shape, the sticks giving a stronger field).

Orders
of magnitude of magnets are listed here.

A typical refrigerator magnet has a field
of 5 mT (milliteslas), with energy density 10 J /m^{3} (to
be compared to the 57 J /m^{3} of the limit electric field
in the air).

Next section :

An exploration of physics by dimensional analysis

Set theory and foundations homepage