The potential is given by $$\Phi(\mathbf{x})=\int_V d^3x' G_D(\mathbf{x},\mathbf{x'})\rho(\mathbf{x'})-\frac{1}{4\pi}\oint_S d^2x'\frac{\partial G_D(\mathbf{x},\mathbf{x'})}{\partial n'}\Phi(\mathbf{x'})$$ Is there a physical interpretation of the surface term? The only one I can think of is that it is the surface integral of $\Phi(\mathbf{x'})$ times $\sigma(\mathbf{x'})$, where $\sigma(\mathbf{x'})$ is the surface charge induced by the presence of a unit point charge at $x$, with the surface being grounded. But why it should be of this form?

  • $\begingroup$ Look at comment in Jackson 3rd Edition page no 37. It clearly explains why this term is electric potential due to a surface charge density. $\endgroup$ – user7757 May 25 '14 at 13:44

Your intuition regarding the surface term is right. Beyond that it is just plain details and I give a derivation below.

Consider the identity (called Green's second identity) $$ \int_V \left(\phi \nabla^2 \psi - \psi \nabla^2 \phi\right)\text{d}V' = \oint_S \left(\phi \frac{\partial\psi}{\partial n} - \psi \frac{\partial \phi}{\partial n} \right) \text{d}S' $$ In this equation, set $\phi(\mathbf{x}')=\Phi(\mathbf{x'})$ and $\psi(\mathbf{x}')=G(\mathbf{x},\mathbf{x}')$ and use $\nabla^2 \Phi(\mathbf{x}) =-4\pi \rho(\mathbf{x})$ and $\nabla'^2 G(\mathbf{x},\mathbf{x}') =-4\pi \delta(\mathbf{x}-\mathbf{x}')$ along with the boundary conditions i..e, $\Phi=\Phi_0$ and $G(\mathbf{x},\mathbf{x}')=0$ for $\mathbf{x}'\in S$. If I not made some trivial errors, you will then reproduce the required relation.

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