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Electrostatic fields have a unit $\frac{V}{m}$ (volts per meter) and the mechanical force fields (including gravitational field) have a unit $\frac{m}{s^2}$. If we exclude gravitational force from this question, as we know all mechanical forces are basically the electromagnetic forces. And using the theory of relativity we can sum up that an electromagnetic field can be viewed purely as either an electric field or a magnetic field by switching to the appropriate frame of reference. Thus we should be able to express the unit of mechanical force fields $\frac{m}{s^2}$ in the unit of electric fields $\frac{V}{m}$. What is the conversion factor then?

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  • $\begingroup$ I don't entirely agree with your units. You say that mechanical force fields have units $\frac m{s^2}$, but this is the unit of acceleration. The real unit of a force is $\frac {kg\cdot m}{s^2}$. $\endgroup$ Jan 6, 2016 at 9:04
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    $\begingroup$ @MikaelFremling The field strength is force per charge. E.g., $E=F/q$. Similarly, a force field would be defined as $F/m$. $\endgroup$ Jan 6, 2016 at 12:42
  • $\begingroup$ @MikaelFremling Field is not force. It is either force per unit mass or charge. $\endgroup$ Jan 9, 2016 at 16:56

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Since I can't think of a good example of a mechanical force field aside from gravity, I'll look at the slightly different approach of comparing the electric field to gas pressure, which is just force per unit area (and dimensionally equivalent to energy density). So the units of pressure are $\frac{kg}{ms^{2}}$ (which is your units of force field divided by area).

The electric field has units of $\frac{V}{m} = \frac{J}{Cm} = \frac{kg.m^{2}}{Cms^{2}} = \frac{kg.m}{Cs^{2}}$ ($C$ is Coulombs, the unit of electric charge)

So what do you have to multiply the electric field by to turn it into a pressure? Something with the units of $\frac{C}{m^{2}}$ - something with the dimensions of charge per unit area.

So if you want to look at gas pressure at the molecular or atomic level, you are looking at the electric field generated by the gas and seeing how that interacts with the density of charge in the container holding the gas - all at extremely short ranges for the gas molecules that are actually very close to the surface of the container. That is what determines how the atoms in the container react to the molecules in the gas (or at the macroscopic level, how the container material reacts to the gas pressure).

This isn't a big surprise: if you want to look at the scale where everything is electromagnetic forces rather than mechanical forces, then you are looking at the scale where everything is interactions between charges.

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The premise "an electromagnetic field can be viewed purely as either an electric field or a magnetic field by switching to the appropriate frame of reference" is false. In general, if you have a point in spacetime, there is not necessarily any reference frame in which B=0 at that point, and likewise there is not necessarily any reference frame in which E=0 at that point.

The premise "we should be able to express the unit of mechanical force fields (m/s^2) in the unit of electric fields (V/m)" is also not true. I don't understand why you think it is true, even after reading the question a few times.

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  • $\begingroup$ Is there a short proof to the statement "an electromagnetic field can be viewed purely as either an electric field or a magnetic field by switching to the appropriate frame of reference is false"? Or an example where there's a field that not reducible to either an E or B field for any observer. Would taking an EM plane wave work? $\endgroup$ Jan 2, 2016 at 22:56
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    $\begingroup$ @no_choice99 When the electric field is very large and the magnetic field is very small then every frame sees a nonzero electric field. When the magnetic field is very large and the electric field is small then every frame sees a nonzero magnetic field. If the relative size is perfectly matched, then every frame will see two nonzero fields. Look up the scalar invariants of the electromagnetic field. $\endgroup$
    – Timaeus
    Jan 3, 2016 at 1:21
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    $\begingroup$ @Timaeus, I see. I knew about some invariants such as |E|^2-|B|^2. If both E and B are equal say like in a plane wave then this invariant is worth 0 and I see no way to see only either an E or B field. So that would answer to the question I asked to Steve B. Does this sound correct? $\endgroup$ Jan 3, 2016 at 1:31
  • $\begingroup$ I had this doubt after watching this video which gave me the impression that magnetic fields can be viewed as electric fields if we choose the appropriate reference frame. youtube.com/watch?v=1TKSfAkWWN0 $\endgroup$ Jan 4, 2016 at 17:25
  • $\begingroup$ @ThePhysicist Your view is common, and it's based on descriptions whose intent is to explain the presence of a real, measurable force where you might naively expect there to be none. Those explanations do not show that one or the other field can disappear or be irrelevant. Your video does not state the facts correctly (there, and in at least one other place, so listen with caution to this guy). See this question and its answers for a good discussion. $\endgroup$
    – garyp
    Jan 5, 2016 at 14:25
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This is a question on choosing the constant of proportionality($K$) in Coulomb's Law. It's worthwhile to mention that coulombs law(expression for electrical force) was first discovered by measuring the mechanical force using a torsion balance.

Electric field and Electric force have different units(clarification to OP). In the terminology used in the question,

Electric field $$E = K \frac{Q}{r^2} ~~~~~~\text{ have the unit } ~~~~\frac{V}{m}$$

Electric force $$F = K \frac{Qq}{r^2} ~~~\text{have the unit }~~ \frac{VC}{m}$$ Now,

$$ 1 ~\frac{VC}{m} = 1~~Newton = 1 \frac{kg~m}{s^2}, \text{when}~ K = \frac{1}{4\pi\epsilon_0}$$

You could define $K$ differently to have a new set of units. For e.g. $K=1$ in gauss units.

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  • $\begingroup$ your answer still does not help in expressing electric field units in terms of mechanical fields or vice versa. As on one side you'll have C and on the other side you'll have V. It is a question that asks about the relationship between charge and mass of a particle. Considering the quantum of charge to be e (electron), we see that if there is a charge difference, there must be a mass difference. However the reverse is not true. $\endgroup$ Jan 9, 2016 at 17:09
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I think we need to start by being clear about a "mechanical" force. There are only four fundamental forces that we know of (gravity, electromagnetic, strong and weak). All of them apply changes to momentum (by their very nature), and thus could be considered "mechanical".

If by mechanical you mean the resultant force, $F = \sum_i\frac{d\rho_i}{dt}=\sum_i m_i a_i$ (last step only true when mass is constant with respect to time), then please consider the case where the mechanical ($F_{mech}$) and electric forces ($F_{elec}$) are the same $$ F_{mech} = F_{elec} $$ multiply left-hand side by $m/m$ and the right-hand side by $q/q$ $$ \frac{F_{mech}}{m}m = \frac{F_{elec}}{q}q $$ we see that the conversion factor from mechanical to electric field strength is $q/m$. Thus, the unit conversion factor is $C/kg$.

This can also be seen using the S.I. definition of Voltage, $V = kg \ m^2 \ s^{-3} \ A^{-1}$. Using the definition of Current (charge per time), $A = Cs^{-1}$ and plug into your units for electric field strength $$ \frac{V}{m} = \frac{kg \ m^2 \ s^{-3} \ C^{-1} s }{m} = kg \ m \ s^{-2} \ C^{-1} = (\frac{m}{s^{2}}) \frac{kg}{C} $$ So $$ \frac{V}{m} (\frac{C}{kg}) = \frac{m}{s^{2}} $$

I hope this answers your question, although I fear there may also be confusion over field strength. All forces are defined between pairs of particles, so to obtain the strength of the force from one particle we divide out the dependence on the other particle. However different forces act on different properties, e.g. for electromagnetic we divide by charge and for gravitational we divide by mass. So it makes sense that the field strengths have different units. You've interpreted $F/m$ as a general mechanical field strength, but this only true for gravity (because gravity acts on mass, and the others don't).

Edit: Sorry for misunderstanding your question, here's my best answer to the relationship between mass and charge.

Currently we don't know of a fundamental relationship between mass and charge, and if there is one it's likely to be one of the last pieces in a Grand Unified Theory (GUT). Our current best model, the standard model, which is a kind of Quantum Field Theory + Special relativity, has successfully unified the electromagnetic and weak forces. However it has some serious problems which Wiki can tell you all about.

(Nutty) Food for thought:

One interpretation of charge from the Dirac equation (QFT + SR) is that anti-matter (opposite charge) is normal matter travelling backwards in time. So maybe charge and time are intimately related, and by unifying QFT with General Relativity (our best description of time and space) we'll find the relationship between mass and charge.

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  • $\begingroup$ I tried exactly the same way as you did. Your answer still does not help in expressing electric field units in terms of mechanical fields or vice versa. As on one side you'll have C and on the other side you'll have V. It is a question that asks about the relationship between charge and mass of a particle. Considering the quantum of charge to be e (electron), we see that if there is a charge difference, there must be a mass difference. However the reverse is not true. $\endgroup$ Jan 9, 2016 at 17:12
  • $\begingroup$ I'm sorry if I misunderstood the question, but I'm trying to argue that electric field units are fundamentally different to mechanical field units; in the same way that you can't express meters in seconds. To help me help you, could you please clarify the following for me? What do you mean by mechanical field? Sorry I don't understand, where is the charge difference? $\endgroup$
    – Judge
    Jan 9, 2016 at 17:33
  • $\begingroup$ The exact relationship between charge and mass is currently unknown, maybe in the future we will have a Grand Unified Theory (GUT) that unifies the electric and mass fields (probably by unifying the photon and Higgs boson). Take a look at the standard model for our best current theory. Note that the Higgs and photon do not interact directly. Am I getting closer to what you're after? $\endgroup$
    – Judge
    Jan 9, 2016 at 17:56
  • $\begingroup$ By a mechanical force field all I mean is force per unit mass where as for electric field it is force per unit charge. If two particles have different charge then there must be difference in number of electrons they are carrying individually however for different mass it is not true. @judge got my point. I realize that so far there is no unity between mass and charge. But thanks for trying to help me. $\endgroup$ Jan 11, 2016 at 3:57
  • $\begingroup$ Have edited my post so that includes my answer. Caveats: Quarks have fractional charge, so it may to be 100% correct to say that the electric charge is the quantum of charge. Although quarks are confined into combinations that have an integer amount of electric charge. There is ambiguity in "force field". Every force can be thought of as a field. The "field strength" is the force/mass for gravity and the force/charge for electromagnetic. For constant mass, the force/mass is the acceleration due to the field. $\endgroup$
    – Judge
    Jan 11, 2016 at 10:02

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