# Boundary condition for a floating electrostatic potential

I have a (probably) simple question regarding boundary conditions. In electrostatic simulations, the relevant Maxwell equation is $\nabla \cdot \mathbf{D}=\rho$ where $\mathbf{E}=-\nabla V$, and $\mathbf{D}$ is related to $\mathbf{E}$ through any applicable constitutive relations. When a floating-potential metal object is embedded inside a domain, I've read that the relevant boundary condition for the metal surface is $$\int_{\partial\Omega}\mathbf{D}\cdot\mathbf{n}dS=0.$$ For example, in this picture the red square is at potential 1V, the blue square is at 0V and the bottom edge is an electrically-isolated metal surface:

Is there a simple explanation for why the integral boundary condition indeed describes a floating equipotential metal surface? The source I found it in (COMSOL's documentation) does not provide a derivation. Apologies if this is intro E/M stuff that I've just forgotten. I keep thinking of using the divergence theorem, but the boundary $\partial\Omega$ does not necessarily enclose a space over which a volume integration may be performed.

• Which COMSOL version is this? – Gotaquestion Nov 16 '13 at 16:32
• I've always taken the boundary condition at the surface of a conductor to be that the tangential components of the electric field are all zero. – Jerry Schirmer Nov 16 '13 at 17:52
• @JerrySchirmer: That makes sense and was what I originally thought, since the electric potential gradient along the direction of the surface should vanish. But apparently that's not the condition that numerical solvers use? – DumpsterDoofus Nov 16 '13 at 18:36
• @Gotaquestion: It's 4.3, although I don't think the version number makes much difference for electrostatics. – DumpsterDoofus Nov 16 '13 at 18:37
• I completely agree with you but I haven't seen an integral formulation in COMSOL documentation. As far as I can remember they always use differential formulation for standard modules. I will have a look when I go to work @DumpsterDoofus – Gotaquestion Nov 17 '13 at 1:28

Yes there is a simple explanation. Think of a metallic sphere that is not connected to anything (i.e. floating), move this conducting sphere close to a positive electrode. Now what happens is, the negative charges are induced on the surface of the sphere closest to the electrode; those negative charges leave positive charges behind. Just like figure (a) here:

The charge that accumulated at the surface because of induction by electrode affects the voltage everywhere. In electrostatics, the method of image is used to include the contribution of induced surface charge to the total voltage everywhere.

The only parameter that one needs to describe a floating potential conducting sphere is the initial charge. In the figure above the total charge is zero. If the sphere was initially charged by a positive or negative charge, the induced surface charge will be affected depending on the initial charge magnitude and polarity.

Now speaking of mathematics, the electric flux is related to the volume charge density by Gauss law:

In the previous equation, D is the electric flux and rho is the volumetric charge density and Q0 is the total charge in the volume. The previous equation basically states that the charge enclosed within a volume generates an electric flux that if integrated on the surface of the enclosing volume gives a constant value equal to the charge enclosed. In conductors it is known that there is no volumetric charge, instead there is only surface charge, so for the conducting sphere case Gauss law becomes:

The difference between the left hand side and the right hand side is that LHS is evaluated anywhere in space where r (the radial distance) > R (the radius of the sphere). RHS is only evaluated at the surface of the sphere where surface charge is located.

Now in your example, first you assumed zero initial charge (that is why the right hand side of your condition is zero); second you only care about the part of conductor that faces the electrodes, that is why your surface integral is open rather than closed as Gauss law dictates.

To confirm this explanation try to plot the surface charge along the edge on which you defined a floating potential. I replicated your model, the voltage everywhere is:

I plotted the surface charge density along the floating potential boundary:

You see, a negative charge was induced close to the positive electrode leaving an identical positive charge behind (close to grounded electrode), which is what happened in figure (a) above. If you integrate the surface charge density along the boundary the result is zero (there is no initial charge). If you used non-zero initial charge, the integration result will be equal to that value. Keep in mind the accuracy of the solution depends on the resolution of the mesh, so the value in case of zero charge is never equal to zero exactly because of numerical error.

• Thanks a lot for writing this. The line graph of induced charge in the metal plane definitely makes sense. I guess my sticking point is a bit more math-related than physics-related. I'm aware of the fact that all the charge on a conductor is surface charge which means $Q=\int_{\partial\Omega}\rho_sdS=\int_{\partial\Omega}\mathbf{n}\cdot\mathbf{S}{dS}$ as you mentioned. – DumpsterDoofus Nov 18 '13 at 14:24