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?
[Physics] Why is there only a drop in potential energy when charges flow through a resistor
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Electric potential is a potential energy just like gravitational potential energy or indeed any other form of potential energy. Specifically, moving one coulomb of charge through an electrical potential of one volt produces (or requires) 1 joule of energy. From your question I guess you're basically happy with this, so the question is really how this energy is dissipated i.e. what happens to that 1 joule of energy?
When you apply a voltage to the conductor you produce a force on the conduction electrons so they accelerate - the potential energy is turned into kinetic energy of the electrons. However conductors are made up from a crystal lattice of atoms/molecules that is randomly vibrating due to thermal energy, and there is a probability that the moving electrons will scatter off this lattice and transfer energy to it. So the electron is slowed down and the magnitude of the lattice vibrations is increased. Increased lattice vibrations mean the conductor is hotter, so the kinetic energy of the electrons has been transferred into thermal energy in the conductor.
And that's what happens to the 1 joule of energy. It's transferred to the conductor and ends up as heat.
Some related issues you might want to look into further: when you cool a conductor you reduce the magnitude of the lattice vibrations and you make it less likely the electron will scatter off the lattice. That's why resistance (usually) decreases with decreasing temperature. The superconducting transition prevents electrons from scattering off the lattice, so they can't transfer energy to it and that's why superconductors have a resistance of zero.
It is best not to think in terms of fields if you are doing circuits. The primary quantities of interest in circuits are voltage and current. The E field is the spatial derivative of the voltage, eg the units of the E field is V/m. However, in a circuit the position is abstracted away. There is no indication of the length of any wire or resistor or battery. So there is no way to obtain information about the E fields given only information about the circuit theory representation of the circuit.
Inside a resistor the E field should be more or less uniform, with a value approximately equal to the voltage across it divided by its length.
Inside a capacitor the E field should be more or less uniform between the plates and zero within the plates.
Inside a battery the E field should be quite strong right at the electrode where the electrochemical reaction makes the voltage difference occur over molecular-scale distances. The E-field will be substantially lower elsewhere, including approximately 0 within the metallic conductor and relatively close to 0 within the electrolyte. Calling it uniform is not accurate.
Best Answer
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.
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.