My question is: is there a simple and truly general equation for the resistance between two electrical equipotential surfaces?. Obviously, if so, what is it, and if not, why? It would be very difficult to solve, granted, but I just want to see a calculus equation that is fully descriptive. I have two frameworks under which this could be entertained, I'll write those out and then explain the motivation.

To start with, we need propose that the volume separating the two surfaces has a volumetric resistivity, $\rho$ in units of $(\Omega m)$.

Single Volume Framework

We can limit the discussion to a defined volume, then the surfaces reside in that volume or on the surface of it. This volume may have a constant resistivity $\rho$ while everywhere outside the volume is completely electrically insulating.

Infinite Volume Framework

An alternative to the above approach that might make the task more or less difficult would be to replace a constant resistivity with a spatial dependence $\rho(\vec{r})$ and no longer require a boundary condition. In that case we only have 3 mathematical inputs to the problem, which is the resistivity defined for all $\vec{r}$ and a definition of the two surfaces, $S_1$ and $S_2$.

Known Algebraic Analogs

The basic algebraic formulation that I find insufficient is:

$$R = \rho \frac{\ell}{A}$$

Where $l$ is the length of the restive material that is any shape which has translation symmetry over that length, and $A$ is the cross-sectional area. Obviously, this is a rather simple equation that won't apply to more complicated geometry. Even more sophisticated academic sources seem to give equations that fall short of what I'm asking. For example:

$$R = \rho \int_0^l \frac{1}{A(x)} dx$$

I think it's obvious that an equation such as this is built upon a myriad of assumptions. For a thought experiment, imagine that the area starts out as very small and then pans out to very large quickly. Well, accounting for the larger area in the above sense underestimates the resistance, because the charge has to diffuse out perpendicular to the average direction of flow as well as parallel to it.

I have some reasons to suspect this might actually be rather difficult. A big reason is that all the approaches I'm familiar with require the flow paths to be established beforehand, which can't be done for what I'm asking. So maybe this will result in two interconnected calculus equations.


I had an interest in Squishy Circuits, and it occurred to me that I can't quickly and simply write down the equation for resistance between two points. The unique thing about Squishy Circuits is that it calls for two types of dough, one that conducts and one that is mostly insulating. However, the recipes aren't perfect and because of that, the young children who play with these circuits regularly encounter the limits of conductor and insulator definitions. If you make your conductor dough too long and/or too thin, you will encounter dimming of the light you connect with it. Similarly, a thin insulator layer will lead to a lot of leakage current which also dims the light.

  • $\begingroup$ Of course there is. With a little thought you should be able to deduce it from the series and parallel rules for discrete resistors. An important word here is "resistivity". $\endgroup$ Mar 4, 2012 at 23:19
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    $\begingroup$ BTW--I think that you have written "conductivity" where you mean "resistivity". They differ by an inverse. $\endgroup$ Mar 4, 2012 at 23:30
  • $\begingroup$ @dmckee Does your 'series-parallel' method take care of the fact that electrons will diffuse out and encounter more resistance during this diffusion if the area increases? It doesn't look like it. There ought to be an additional heat loss. $\endgroup$ Mar 5, 2012 at 0:55
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    $\begingroup$ @Manishearth It works fine, but it occurs to me that you need to have solved the electric field before you can begin because the areas must always be computed along the local direction of flow (i.e. in the direction of the local electric field); this is really the only assumption underling the relations in the post and series part is actually expressed in the relations that Alan gives. $\endgroup$ Mar 5, 2012 at 2:22
  • $\begingroup$ @dmckee yes, exactly. I doubt that series-parallel can handle that. I think $J=\sigma E$ would do the trick. But to calculate E, well, O_o. Now I'm horribly confused :p $\endgroup$ Mar 5, 2012 at 2:29

3 Answers 3


I will do the case where the material is homogeneous and isotropic, $\rho = \sigma^{-1}$ is a constant proportional to the identity matrix. We are interested in the steady state, where none of our variables depend on time.

We have $\nabla \times E = 0$ from Faradays law and, $\nabla\cdot J = 0$ from the equation of continuity, where $J$ is the current density. Ohm's law tell us that $J=\sigma E$. Taking the divergence of Ohm's Law we get $\nabla \cdot E = 0$. Therefore in steady state $E$ must be a divergence-free, curl-free function.

This means that the potential $\phi$ ($\nabla\phi =E$) obeys Laplace's equation,

$\nabla^2 \phi = 0$.

To solve this we need the appropriate boundary conditions which are as follows. At the boundary where your resistor connects to a lead the potential $\phi$ must be the same value as the potential on the lead. At the boundary of your resistor not connected to a lead you can't have current flowing out, so the appropriate condition is $\nabla\phi\cdot n = 0$, where $n$ is the surface normal. This is sufficient to determine a single solution to Laplace's equation.

Laplace's equation is a very nice and friendly equation and there is a lot of material on numeric and analytic solutions available, although the mixed boundary conditions will be annoying.

Once you have your solution to Laplace's equation you need the total current. To get that, pick any cross sectional surface $S$ of your resistor, and using $J = \sigma E = \sigma\nabla\phi$, we get the total current I is equal to

$I = \sigma\int dS\cdot\nabla\phi$.

Then you can use $R = V/I$ to get an effective resistance. The case with a non homogeneous or non isotropic material is similar, you just end up with a different equation from Laplace's equation, which may be a little more annoying to solve.

I can't imagine a dough based system will need this level of precision though :). For anything dough based probably just assuming its a cylinder and using $R = \rho L /A$ will get you close enough no? I always thought of dough physics as really being an order of magnitude sort of game, like astronomy.

  • $\begingroup$ Besides the fixed potential as a boundary condition, you can also have a fixed current as a boundary condition ($\nabla\psi \cdot n=J_{in}/\sigma$). If all your boundaries are given by a gradient, you need an additional reference point for zero potential at some location. $\endgroup$
    – Bernhard
    Mar 5, 2012 at 7:19
  • $\begingroup$ One followup question: What if there is a boundary between different materials? What can we do for that? To pick the low-hanging fruit, obviously the potential and current is continuous. so maybe $\sigma_1 \nabla \phi_1=\sigma_2 \nabla \phi_2$ and $\phi_1=\phi_2$ where 1 and 2 are either sides of the boundary, but I'm not sure if that's sufficient. The isotropic requirement is something I had not even considered before, but I take it that would make $\rho$ into a more complicated vector quantity? And yes, dough is "messy" but there's beauty in these mathematics. $\endgroup$ Mar 5, 2012 at 7:20
  • $\begingroup$ @AlanSE These continuity requirements will be sufficient imo. For anisotropic materials, this conductivity may even become tensorial. But that is probably not necessary here. $\endgroup$
    – Bernhard
    Mar 5, 2012 at 7:33
  • $\begingroup$ @AlanSE: Yes those boundary conditions are correct. In an anistropic crystal Ohm's law is $J_i =\sum_j \sigma_{ij}E_j$, so $\sigma$ is a matrix. $\endgroup$ Mar 5, 2012 at 17:14
  • $\begingroup$ For the lesser mathematical minds, would you care to explain with an example, say resistance along the diameter of a sphere? $\endgroup$
    – Severo Raz
    Nov 18, 2014 at 5:44

To simplify, let's think of a prismatic volume, as if a planar figure (in $xy$ plane, area $A$) had been extruded in $z$ direction. Let's assume that conductivity ($\sigma$) is dependent on $x$ and $y$, but not on $z$. And let $L$ be the prism length in $z$ direction. We want to compute total resistance between both ($xy$) parallel faces.

If we take a small prism with bases at main prism bases (so the little prism is of same length $L$ as the main one, but its base is of a small area ${\rm d}A$), its resistance cannot be called ${\rm d}R$ (it's not an infinitesimal), instead it tends to infinity ($R_{{\rm d} A} = \rho \frac{L}{{\rm d} A} \rightarrow \infty$). So I propose to consider the inverse magnitude, the conductance, $G$.

If the body were homogeneous, the total resistance would be $R = \rho L/A$, so the total conductance would be $G = \sigma A/L$. You know $\rho$ and $\sigma$ are just inverse quantities. So for the small prism, the conductance is: $${\rm d}G = \sigma \,\frac{{\rm d}A}{\it L}\,.$$ To compute total $G$, we just integrate in $x$ and $y$: $$G = \int {\rm d}G = \int \sigma \frac{{\rm d}A}{L} = \frac{1}{L}\int \sigma\, {\rm d}A\,.$$

Once we have that integral, we can just take the inverse to compute the total resistance: $$R = \frac{1}{G} = \frac{L}{\int \sigma\, {\rm d}A}\,.$$

The above procedure is, of course, only valid assuming a uniform electric field between parallel ($xy$) faces (i.e., $\vec E$ has the $z$ direction). Otherwise, the 'harder' orthodox method (previous answer: computing $\phi$, then electric field ($\vec E = -\nabla \phi$), then current density (from Ohm's law: $\vec J = \sigma \vec E$) and finally current intensity ($I = \int J_z {\rm d} A$)) would be necessary.


There is not such an integral because resistance is a property of whole circuits that may be defined in isolation only in an approximate way. The closest I have come to an integral expression is eq. 63 in https://www.academia.edu/1841457/The_Notion_of_Electrical_Resistance

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    $\begingroup$ Could you copy equation 63 from the reference into your post here? That way, people can make sense of the answer without any external resources. Some more explanation would also help. $\endgroup$
    – David Z
    Jun 9, 2016 at 20:39

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