I'm having trouble understanding how this can be possible. Obviously, I'm missing something but I'm not sure what it is.

The amount of voltage at a point depends on the electrical potential energy at that point, which depends on the density of charge (electrons). So how can the voltage drop if the density of charge can't change, since that would cause a change in current.

I've tried thinking about it two ways and run into problems each time…

  1. The charge density does increases but the speed decreases causing the same current. But if this were the case, and the density did increase in the resistor, then the resistor would have a higher potential than the wire and charge would stop flowing.

  2. The charge density decreases across the resistor and the resistor acts like a funnel, creating a higher charge density at one end than the other. But the difference in charge density is counteracted by an increase in speed, allowing the current to remain constant. However we know resistance slows down charge, it doesn't speed it up.

  • 1
    \$\begingroup\$ Can you be more specific what is the question? An ideal constant current source would push constant current through any load which means any voltage drop is possible but Ohm's law will define how current, voltage and resistance relate to each other. \$\endgroup\$
    – Justme
    Nov 25, 2021 at 23:13
  • \$\begingroup\$ Voltage is how much potential energy each charge has. It does not depend on the "density" of charges. It should not be surprising that a charge giving off some of it's energy as heat causes the amount of energy it retains to decrease. \$\endgroup\$ Nov 25, 2021 at 23:16

3 Answers 3


Maybe you're confusing electric field strength with electrical potential energy. It's true that the two ends of a resistor can have different potentials, endowing the charges at one end with more potential energy than charges at the other, but the electric field inside the resistor, between those ends, is uniform, it's strength being equal at all points.

There may be a voltage source somewhere, like a battery or capacitor, that has a non-uniform distribution of charges, where you may associate potential with charge density, but the field resulting from that potential difference is not confined to the battery. Any resistor across that source is effectively a uniform distribution of charges within a uniform field, with equal force on each charge in the conductive path, and no charges grouped up anywhere inside the resistor.

Potential is a measure of how much work can be done by a charge in travelling from one point to another, consistent with the concept of work done equalling force times distance. The force on each charge in the resistor is the same throughout the resistor, because the field strength at all points is the same, but some charges will have to move further to get to the end with the lower potential. Therefore some charges are able to do more work than others (those nearer the high potential end), because they have to travel further, dispite all being pushed by the same amount of force.

You can't associate potential with charge density, in the same way you can't say that there are more blades of grass per square metre at the top of a hill than at the bottom because grass at the top has more gravitational potential energy. Soccer balls rolling down hill are not moving because there are more balls per square meter at the top of the hill, they are moving because they find themselves in a gravitational field.

You can say that somewhere there may be a charge density imbalance giving rise to a potential difference, but charges in the circuit connected across that potential difference are merely passengers its field.


Electric field strength is a separate thing from electric charge. Charge is what electrons have, electric fields are what is produced by a battery or generator. The electric field pushes the electrons around the circuit.

If you separate the two concepts, the problem goes away.


In a steady state situation, in the interior of a uniform conductor, the charge density is 0. This does not mean that the current density is 0. It means that the positive and negative charges are present in equal amounts (at above atomic level scales).This result is a consequence of the microscopic version of Ohm's Law and Maxwell's (Heaviside's) equations.

What applies to a uniform conductor, applies also to a uniform resistor. In steady state the charge density is 0.

Where charges accumulate is at the boundaries where the resistivity (or conductivity) of a conductor changes, or where the current density changes due to a change in cross section or a junction in the wires.

It is the accumulates charges in these locations that "generate" the electric field seen in steady state.

However, it is not necessary to calculate these charge densities to calculate the electric field within a conductor/resistor. The electric field is given by the microscopic version of Ohm's Law.

$$\vec{j} = \sigma\vec{E}$$

where \$\vec{j}\$ is the current density, \$\sigma\$ is the conductivity, and \$\vec{E}\$ is the electric field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.