# Equilibrium of electrostatic potential

we often talk about the similarities between the temperature and the electrostatic potential. However, we do not know of exact analogue of the Heat diffusion equation $$\nabla^2{T}=\alpha\frac{\partial T}{\partial t}$$ in electrostatics. Let us consider an example where this can be relevant. Say, we have two dielectric blocks at specific potentials $$V_1$$ and $$V_2$$ (well, this cannot be strictly true, but let us say that the variation of the potential in each of them is much less than their values). Now, we join them together along a common boundary. Naively it appears that the boundary should come to the same potential after some time. Is there a way to estimate how long this may take? There should be some method to deduce this, because for conductors, for example, this is almost instantaneous. So, somewhere the dielectric constant of the bodies will come into the picture...If anyone can shed light on this, that will be very much appreciated. We do find the Telegrapher's equation that involves potential and its higher order temporal derivatives, but I do not think it could be used directly in this context.

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• Why do you think " Naively it appears that the boundary should come to the same potential after some time." if the material is a non conductor it should not. But you may consider whats called contact electricity, you can charge a non conductor by bringing it in close contact with a other, (Wool and plastic or glas for example is often used.) Sep 28, 2021 at 17:27
• @trula: The reason is there is going to be a potential gradient across the boundary - almost like a step. This will facilitate flow of charge, though it may not be very fast, because we are talking about dielectrics. The contact electricity you mentioned most probably involve electric discharging and once that happens, I am almost certain that eventually the boundary of the two bodies come to the same potential. Sep 29, 2021 at 8:12

In the electrostatics case, you can use the constitutive relation $$\vec{D} = \epsilon \vec{E}$$ as well as Gauss' law $$\nabla \cdot \vec{D} = \rho$$ and the definition of the electrostatic potential $$V = -\nabla \vec{E}$$ to arrive to Poisson's equation for the eletrostatic potential:$$-\epsilon\nabla^2 V = \rho$$ To get an idea of the time scale, taking the divergence of Ohm's law gives $$\frac{\partial \rho}{\partial t} = -\frac{\sigma}{\epsilon }\rho$$ So that the free charge decays as $$\rho = \rho_0 e^{-(\sigma/\epsilon )t}.$$
This decay time $$\frac{\epsilon }{\sigma }$$ is in the order of about $$10^{-17}$$ or $$10^{-18}$$ seconds for most metals.