Q:

Capacitance due to accumulation of charge on the interface between resistors

It is well known that the continuity condition for current density necessitates that at a steady state:
$$\frac{\rho}{\epsilon_0}=-\frac{E\cdot \nabla \sigma}{\sigma}$$
Where $\sigma$ is the specific conductivity, $\rho$ is the charge density, $E$ is the electric field vector and $\epsilon_0$ is the permittivity of free space.
Due to the continuity condition being:
$$\nabla\cdot J=\nabla\cdot(\sigma E)=\sigma(\nabla\cdot E) + \nabla\sigma\cdot E = \sigma\cdot\frac{\rho}{\epsilon_0}+\nabla\sigma\cdot E=0$$
This condition suggests that on the interface between two resistors connected in series there is an accumulation of charge. This is also addressed and explained in this question.
All of this makes perfect sense, but raises the question: does this not imply a certain capacitance inherent to the connection of two resistors in series?  
Usually when considering a problem of resistors in series one would simply write $R_{tot}=R_1+R_2$, treat the two as a single resistor and be done with it, seemingly ignoring the capacitance of the interface when (perhaps) solving the circuit's RCL differential equation.
Is my thought process completely off or does this capacitance exist and usually assumed negligible? If so, are there important cases where this effect cannot be neglected?

A:

There are “stray” or “parasitic” capacitances and inductances associated with resistors. There are manufacturer write ups with more specifics. 
The capacitance comes from the connections, but also from the resistors themselves. Different types are better or worse. Surface mount is generally better than through-hole; film resistors are better than carbon. 
These effects are only an issue at higher frequencies. Good components and mechanical design can go to 100’s of MHz with through-hole and low GHz with surface mount. 
Generally, the build-up of charge at the ends of the resistor (to create an electric field in the resistor) is going to be a negligible effect.  For example, a $1/4$ watt carbon resistor is roughly a 2mm diameter tube 6mm long. We can model that as a parallel plate capacitor with
$ C = k \epsilon_0 A/d = 1\times (8.8.5 \times 10^{-12}) (0.001^2 \pi ) / 0.006 = 5\times 10^{-15} = 0.005 \rm{pf}$
That's much smaller than other effects. 

