Extremely confused about Ohms law Consider there are 2 bodies placed at different locations in space. Now I maintain them at a potential of V and -V respectively. If a conducting wire is placed connecting each other, current starts to pass. 
The power dissipated is described by $$P=I^2R$$ or$$P=V^2/R$$
The questions I have in mind are: 


*

*If there are no other electrical components, all of the electrical energy is converted into heat energy (right? I mean I think it is right conceptually). But if I cut the wire and add a motor, nothing in the equation changed. The Voltage didn't change neither did the Resistance, but the motor works. This implies that the energy loss is not equal to the electrical energy in the first case too. Where am I going wrong?

*And even though I can derive this equation, I've realized that I don't fully understand what the equation says. Is this the power that is transferred when there is a constant voltage?

*If the 2 bodies were to be charged up to potentials V and -V, how much energy would be required? As far as I know, I would calculate the energy of the final field, and then to add up the lost energy, is it supposed to be $$V(t)^2/R$$ integrated with respect to dt. Where V(t) can be an arbitrary function. The constraint is that the initial and final voltage differences should be 0 and 2V. Is this right?
Thanks
 A: Forget about resistance.
If you have a potential difference of $V$ across two points and there is a current of $I$ flowing between them then you can say the electrical power is $IV$.  
It might be that all of that electrical power is converted to heat (resistor) or heat and mechanical energy (electric motor) or heat and light (light bulb) or heat and chemical energy (charging a battery) etc.

But if I cut the wire and add a motor, nothing in the equation changed. 

That is not true.
If you have a resistor and a rotating motor coil in the circuit the rotating coil produces a back emf and so affects the current flowing in the circuit.
Have a look at this link which gives an indication as to what happens when a dc motor is in circuit.
A: 2 bodies placed at different locations ... I maintain them at a potential of V and -V respectively
You are making this too complicated and building the wrong mental picture.  It's not about two bodies at V and -V.  V and -V with respect to what?  If you were to discharge these two bodies by letting current flow between them, their relative potential would change.  In effect, you are describing a capacitor.  That's not so useful for understanding Ohm's law.
Instead, think of two points being held at a potential V between them.  This is what a bench power supply does.  There is no need to imagine exotic arrangements of bodies floating in space.
If a conducting wire is placed connecting each other, current starts to pass.
Yes, but a truly perfect wire connected across a truly perfect voltage source is like dividing by zero.  In the real world you can connect a wire across a power supply.  But what then happens depends on the non-ideal characteristics of both the wire and the power supply.
If you want to keep this discussion theoretical, then you simply can't divide by zero connect a wire across a voltage source.
If there are no other electrical components, all of the electrical energy is converted into heat energy
No, the theoretically ideal wire across the ideal voltage source can't exist.  See above.
In the real world, the wire is a resistor, although one with rather low resistance.  If you connected it to a beefy enough power supply, lots of current would flow, and lots of power would be dissipated in the wire.  With a good enough power supply and thin enough wire, you can try this yourself.  It's not hard to get really thin wire (like 30 guage for example) to glow and then break.  Be sure to try this with something not flammable under the wire.
This is exactly the principle behind how incandescent lightbulb filaments are made hot enough to emit visible black body radiation.
But if I cut the wire and add a motor, nothing in the equation changed.
Oh yes it has.  The motor has some deliberate and finite resitance.  You can use Ohm's law to calculate the current when the motor is stalled.  There is a voltage across the motor and a current thru it.  That voltage times the current is the electrical power going into the motor.
I don't fully understand what the equation says
Maybe a example will help illustrate the concepts.  There are two relevant equations.  Ohms law:
    V = IR
where V is the voltage across a resistor, I the current thru it, and R its resistance.  Think about what this says.  A resistor is like a constriction in a pipe.  The more flow (current) you try to put thru it, the more pressure (voltage) will build up across it.
The other relevant equation is:
    P = IV
where P is the electrical power going into something with I current thru it and V voltage across it.
By combining these two equations you can get:
    P = I2R = V2/R
Let's do a example, one you can even try for yourself.  Let's say you have a voltage source (power supply) that puts out a constant 12 V.  You connect a 20 Ω resistor across it.  What will happen?
By Ohm's law, you can find the current.  (12 V)/(20 Ω) = 600 mA.
Knowing the voltage and current, you can find the power dissipation.  (12 V)(600 mA) = 7.2 W.  Notice that you could have figured the power dissipation directly from any two of current, voltage, and resistance.  We already did current and voltage.
By current and resistance, the power is (600 mA)2(20 Ω) = 7.2 W.
By voltage and resistance, the power is (12 V)2/(20 Ω) = 7.2 W.
