How would I explain Ohm's Law in terms of Electrical Fields and Force? In terms of current, resistance, and voltage, it's easy: Ohm's Law is the relationship between current, voltage, and resistance of a circuit. Boom, simple as that. How could I put this in terms of $E$ and $F$? I can sort of see a way to do it by relating the formulas $E=F/q$ and $I=q/t$ to Ohm's Law, $V=IR$, but I'm not entirely sure how I could explain this in words.
 A: There are a number of ways you can examine the law in a microscopic view. One of them is this:
An applied voltage creates an electric field, which superimposes a small drift velocity on the free electrons in a metal conductor. This drift velocity is way smaller than the speed of transmission in a conductor.
Now, the basic relations are: 
$$
I=\frac VR\\
J=I/A//
R=\frac{\rho l}A
$$
From the above, we can get:
$$
J=\frac V{RA}=\frac V{\rho l}=\frac {El}{\rho l}=\frac E{\rho}=E\sigma
$$
These relations can help you put the equation in terms of E, F or whatever else it is you want.
A: It is possible to arrive at the expression for Ohm's law by using a simple classical model. The simplest treatment I've seen of this happened to be on an optics book: Pedrotti and Pedrotti's "Introduction to Optics", Chapter "Optical Properties of Matter", paragraph "Conduction current in a Metal".
Basically free electrons in metals can be thought to obey the differential equation
m v' + m gamma v = -q E

where gamma is a frictional constant, m and q the mass and (absolute value of the) charge of the electron. In the above equation v and E are vectors.
EDIT - Let me try to explain how this is related with the OP question:
In this simple classical model, the force on the electrons is due to the electric field (-q E) and to a viscous 'resistance' proportional to the velocity (and opposed to it in direction, hence - m gamma v). The above equation is just plain old F = m a.
This is the relation between F and E:
m a = F = -q E - m gamma v

It turns out that the viscous drag is responsible for the simple proportionality between current density and electric field that expresses Ohm's law at a microscopic level. 
We can see this by expressing v (vector velocity) in term of the current density j = -q N v
You get an equation in j
j' + gamma j = (N q^2)/m E

whose solution in case of an harmonic Electric field E = Eo Exp[-I w t] can be expressed in term of phasors
j =  (N q^2)/(m (gamma - I w) E

In the static case (w=0) this reduces to
j = (N q^2)/(m gamma) E

or, as you might have surmised
j = sigma E

where sigma is the conductivity and is a constant as long as gamma is a constant. Hence the constitutive equation is linear.
Does this mean that we have 'demonstrated' Ohm's law - turning it into Ohm's theorem? No, the 'mojo of experimentalness' has simply shifted from R to rho to sigma to gamma. 
And it can be pushed further, if we want: IIRC, a more thorough analysis could show how to arrive at the same microscopic form of Ohm's law starting with the parabolic motion of the electrons in the constant field and defining sigma in terms of the mean time between collisions. In a few days I might find the time to look for my notes and expand this.
A: Imagine at the face of the resistor that $N$ electrons each with charge $q$ are collected and move along it with a constant average drift velocity $v$. So in time $dt$ a charge $dQ$ of $N q vdt$ moves past any point, that is, the current $I = \frac {dQ}{dt} = N q v$.
Start with Peltio’s first equation above assuming that $v’ = 0$ (a reasonable approximation), so that
$v = - q E / (m*gamma)$, where m is the mass of the electron.
So $I$ and $v$ are shown to be proportional. Now we finish the argument by showing that also the voltage $V$ across the resistor and v are proportional as follows:
The force $F$ on a single electron $F$ is $q E$, and the work $W$ to move it along a distance $s$, the length of the resistor, is $F \cdot s$. So $W = q E \cdot s$ = $- m*gamma*v*s$. So the voltage $V = \frac{W}{q}$$ = - m*gamma*v*s$$ = -\frac{m*gamma*s*I} {(N q)}$. 
Finally,
$V = I [\frac{m*gamma*s}{N (-q)^2}] = I R$, where R is the bracketed expression.
A: Let's start with $E=Fq$, like you have up top. Rearranging to solve for q, we have $q=E/F$.
We know that a change in charge creates a current, $dq/dt=I$, so substituting $\frac{d}{dt}\frac{E}{F}$ for $\frac{dq}{dt}$, we now have,
$\frac{d}{dt}\frac{E}{F} = I = V/R$
You could also use the relationship $E=-\bigtriangledown{V}$ to obtain a relation.
A: There is no way to derive Ohm's "law" from simple definitions of electric field and current. You have to understand the dissipative mechanisms at work in a system to validate the notion of Ohm's law. Sam29's answer introduces these dissipations through the concept of the resistivity $\rho$ or its reciprocal $\sigma$ of a material. The effectiveness of $\rho$, $\sigma$ and of Ohm's law itself in characterizing the dissipative mechanisms in charge flow is a wholly experimentally established fact.
The resistance $R$ measures the work that must be done on a conductor or conductive system to push charge through it. Microscopically, the charges are thrust by the electric field and, were there no other resistance, these charges would accelerate rather than seem to drift at a constant mean velocity. But they crash into the lattice of the conductor inelastically, transferring some of the energy that the electric field has bestowed on them into heat energy of the vibrating lattice. So they undergo a stop-start, zigzag motion that reaches a constant velocity steady state. It is this steady state that allows Ohm's law to be meaningful.
A: Ohms law is not a "law" as much as it is an excellent approximation of an unavoidable material property. Imagine a charge in vacuum with an contain electric field along z. This charge will continue accelerating, therefore the current is unbound. In this example the system has an inductance (current changes with time at constant applied potential), but no resistance.
If you place that charge in an environment where collisions are likely, you'll have a fast particle moving among particles with average velocity zero. The resulting collisions will slow down the charged particle by transferring energy to the material - which mostly becomes heat. When the applied field equals the average drag force on the charged particle from the lattice, you have constant velocity - constant current. 
Ohm's law goes further by suggesting the relationship is linear, which it is not. Usually this is a good approximation, but be careful because as simple a systems as an incandescent lightbulb seriously deviates from this)
