electric circuits - Voltage drop = more electrons on one side of resistor

I have been asking myself this question for a long time now. Suppose we have two resistors in series connected to a voltage source. Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side? Because if an analog voltmeter is connected in parallel to that resistor, current will flow through the voltmeter.

Answer

Simply put, does the voltage drop on each resistor mean that there are more electrons on one side of the resistor compared to the other side?

Yes, the charge density at one end of the resistor must differ from the other if there is a current through.

Consider, for simplicity, a resistive element of length $L$, area $A$ and resistivity $\rho_r$.

Joined to each end are conductors with area $A$ and resistivity $\rho_c$.

Assume a constant current density of magnitude $J$ along the length of the conductors and resistive element.

The electric field, in the direction of $J$, within the conductors and resistive elements is given by $E_c = J \rho_c $ and $E_r = J \rho_r $ respectively.

Then, at the boundary between the conductor and resistive element, the electric field abruptly changes in value by

$$\Delta E = \pm J\left(\rho_r - \rho_c \right) \approx \pm J\rho_r\;,\quad \rho_c \ll \rho_r$$

This implies a charge density at each end of the resistive element.

See, for example:

For another approach, consider that the slope of the electric potential changes abruptly at the interface.

For a steady current and assuming essentially ideal conductors, the electric potential along the conductors is constant but begins changing with distance inside the resistive element (it must since there is potential difference between the ends of the resistor).

Again, this implies an abrupt change in the electric field at the boundary which requires a charge density at the boundary.