How to express a mechanical force field in the units of electric fields? 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?
 A: 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.
A: 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.
A: 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. 
A: 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.
