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The battery in a closed circuit creates a uniform electric field in the wire, which pushes the electrons from the negative to the positive terminal. The electrons clearly have more potential to do work the farther they are from the positive terminal because they have more distance left to cover, so why then does their electric potential energy not decrease uniformly as they travel toward the positive terminal, but is rather decreased suddenly, when they encounter resistors?

My understanding is that energy is transferred from the electrons to heat energy when they pass through a resistor. However, electric potential energy is dependent on position relative to the positive terminal, which means that electric potential energy should not decrease rapidly across a resistor. If electric potential energy is not converted into heat energy, then which type of energy is? I don't believe it could be kinetic energy because current does not decrease through a resistor.

Why is the above reasoning incorrect? I would appreciate it if you could take it apart in detail.

Additionally, I do not understand the physical significance of resistance. In other words, why is the resistance at a point in a circuit defined as the ratio of the potential difference across the point to the amount of charge passing through the point in a second? Why does this quantity specifically represent the resistance of a resistor to the flow of charge?

Finally, I know this may sound silly, but what stops the large concentration of like charges in both terminals (positive and negative) of a battery from repelling each other and therefore separating, causing the terminals to no longer be charged?

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The electric field is not constant around the circuit. It is nearly zero in the wires, and much more than zero inside the resistors.

The electric field inside a resistor is proportional to the current density $I/A$, where $A$ is the cross-sectional area of the resistor. That is, $E = (I/A) \times \rho$, with $\rho$ a constant. This is observed experimentally. More precisely, this is true for an ohmic material at a constant temperature, and commonly used resistors are basically ohmic. $\rho$ is nearly zero for a material like copper, and is much higher for, say, a typical carbon resistor.

Integrating along the length $\ell$ of the resistor, $\int E \mathrm{d}\ell = (I / A) \times \rho \times \ell = I \times (\rho \ell / A)$. Defining the resistance as $R = \rho \ell / A$, $\int E \mathrm{d}\ell = IR$. And since $\int E \mathrm{d}\ell$ is equal to the potential difference $V$ between the ends of the resistor, you have $V=IR$, which is Ohm's law.

Electric potential energy is converted to heat as current goes around the circuit.

The battery causes electrons to flow through it in one direction (opposite the direction of conventional current), by chemical action. If there is enough chemical energy available, it can create a potential of, say, 1.5V between the battery terminals (example is an AA battery). It's a little like having enough strength to compress a spring; you can compress it, but the more you compress it the harder you have to push to compress it even more. As a battery goes flat, it does a poorer job and the potential declines.

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The battery in a closed circuit creates a uniform electric field in the wire, which pushes the electrons from the negative to the positive terminal.

The electric field in a closed circuit will vary depending on the work required per unit charge to move the charge between various points of the circuit. The work required per unit charge to move the charge between two points is the potential difference, or voltage, between the two points. Consequently, the electric field will only be uniform throughout the circuit if the potential difference between all points in the circuit is the same. And that is because the electric field between any two points in the circuit separated by distance $d$ is

$$E=\frac{V}{d}$$

The electrons clearly have more potential to do work the farther they are from the positive terminal because they have more distance left to cover, so why then does their electric potential energy not decrease uniformly as they travel toward the positive terminal, but is rather decreased suddenly, when they encounter resistors?

It appears you are assuming that the wires connecting the resistors in the circuit have zero resistance. Apart from the fact that, with the exception of superconductors, the resistance of wires is never actually zero, if they were zero resistance then no work would be required to move the electrons between any two points in a wire. That, in turn, would mean the potential difference $V$ between any two points in the wire is zero and there would be no loss of electrical potential energy. The only loss of electrical potential energy would occur in the resistors of the circuit.

If electric potential energy is not converted into heat energy, then which type of energy is? I don't believe it could be kinetic energy because current does not decrease through a resistor.

Electric potential energy is dissipated as heat in a resistor, but not directly. To explain:

The electric field produced by the battery exerts a force on the electrons in the resistor. The work the force does on the electrons gives them kinetic energy. Then the electrons collide with the atoms/molecules of the resistor giving up the kinetic energy in the form of heat. Then the electrons again acquire kinetic energy from the electric field which they again give up via collisions. So the electrons alternatively acquire and give up kinetic energy obtained from the potential energy of the field, so that overall, there is no net change in the kinetic energy of the electrons, which is why the drift current of the electrons is constant. The overall effect is the dissipation of electrical potential energy in the form of heat in the resistors.

Additionally, I do not understand the physical significance of resistance. In other words, why is the resistance at a point in a circuit defined as the ratio of the potential difference across the point to the amount of charge passing through the point in a second?

Your understanding of resistance, potential difference, and current is inaccurate.

First, there is no resistance "at a point". Resistance is between points.

Second, there is no potential difference at a point. Potential difference is between points and, as already stated, the potential difference between two points is equal to the work required per unit charge to move the charge between the two points.

Third, current is not the amount of charge passing through a point. Current is defined as the rate of charge transport through a surface.

So your last statement of Ohm's law should read:

The resistance $R$ between two points in a circuit equals the potential difference $V$ between the two points divided by the rate of charge transport through a surface between the two points, which is the current $I$.

Finally, I know this may sound silly, but what stops the large concentration of like charges in both terminals (positive and negative) of a battery from repelling each other and therefore separating, causing the terminals to no longer be charged?

The only thing that stops it is the absence of a circuit connecting the terminals to enable the charges to separate. In other words, the accumulated charge on the terminals have no where to go if there is nothing connected to the battery terminals.

Hope this helps.

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