Why is there only a drop in potential energy when charges flow through a resistor? First, I read that if you move a charge (say a positive charge) against an electric field, its electric potential energy increases because you're doing work to move the charge to that position. I also read that electric potential is the amount of potential energy per charge. To specify my question, when a positive charge leaves the positive terminal of a battery, ( I know the charge is actually negative, but let's assume it is positive) shouldn't its potential energy already be decreasing with each increase in distance you move the charge away from the terminal? Because it's like moving a charge in the direction of its field, so potential energy should decrease, right?  So shouldn't there already be a drop in voltage ( potential difference) before the charges even reach a resistor?
 A: 
So shouldn't there already be a drop in voltage ( potential difference) before the charges even reach a resistor?

Normally we view a battery or a cell as accumulator of charges in a manner that a potential difference is built up when we charge a battery with plates and electrolyte. 
Therefore if charges flow out or current is being drawn at certain voltage one expects depletion in the reservoir but that variation in the voltage is dependent on the amount of energy drawn-if one unit of charge is moved out the variation will not be measurable.

I read that if you move a charge (say a positive charge) against an electric field, its electric potential energy increases because you're doing work to move the charge to that position. I also read that electric potential is the amount of potential energy per charge. 

Now let us take up the view on "Potential Energy" of the electric field-
One should try to visualize the space around a point charge/charge distribution and the points in coordinate  space having a 'field intensity'
which can be measured by a test charge and defined by intensity vector , based on coulomb's law.
The vector field  can also be mapped by a "scalar potential" which is attributed by the amount of work done in moving a test charge from the point in space where the field intensity is absent/zero (naturally at infinite distance) to the point where field potential is being calculated/defined.
Therefore one should try to understand the 'potential ' as a characteristic of the field rather than the particle.The charge field do provide EMF in driving the charges along a circuit in 'current electricity' and the difference of potential is a measure of the current times the resistance in the circuit- if no load is provided then the avalanche of charge flow may occur- and its a short circuit condition -discharging the battery in a short time interval through sparking/generation of heat etc.-which must be avoided.
A: I think that you have missed an assumption that is often used in circuit theory and that is that the connecting leads have a negligible resistance.  
In the real world the leads do have a resistance and so as the (positive) charge moves from the positive terminal of the battery there is a conversion of electric potential energy into heat and the same conversion happens as the charge passes through the resistor and the same again happens as the charge passes through the leads to the negative terminal of the battery.  
However if the resistance of the leads is much less than the resistance of the resistor the conversion in the leads is very much smaller than in the resistor and so can be neglected.
From this it is assumed that there is no (really negligible) potential difference across the leads.
A: Fist, the amplitude of electric field (thus the electric potential energy) in the wire is not uniform. The amplitude depends on the impedance of the place (I don't know the reason). For example, the eletric field is much stronger in resistor, capacitor or inductor. 
Second, if the electric filed is weak, the drop in the potential will also be small. Since the the impedance of wire is quite small, the drop of potential is negligible.
