A Question From Jackson Electrodynamics I have a question regarding Jackson's Classical Electrodynamics. Consider the equation
$$\varphi \left (  x\right )=\tfrac{1}{4\pi\epsilon _{0}} \int_V \frac{\varrho ( x )}{R}d^{3}x+\tfrac{1}{4\pi} \oint_{\partial V} \left(\frac{1}{R}\frac{\partial }{\partial n}\varphi -\varphi \frac{\partial }{\partial n}\frac{1}{R} \right)dS.$$
This expresses the electrostatic potential due to a charge distribution $\varrho$ in a finite volume $V$ with specified boundary conditions. The first integral expresses the charge inside a volume in space. The second integral encodes the boundary conditions on the surface of this volume.
My first question: does this determine the total potential completely inside that volume regardless of any charge that you put or remove outside the volume? Or does the change in the charge distribution outside the volume cause changes in the boundary conditions (the potential and its normal derivative at the surface)?
My second question: I can understan mathematically that the potential outside the volume should be zero but I can't see why this is so physically. Also, the zero potential outside will change if you put some charge outside the volume, right?
 A: So, the first integral:
$\tfrac{1}{4\pi\epsilon _{0}} \int \frac{\varrho ( x )}{R}d^{3}x$
Is your basic potential from a charge distribution equation.  It is the potential from all charges inside the volume.
And the second integral:
$\tfrac{1}{4\pi} \oint \left(\frac{1}{R}\frac{\partial }{\partial n}\varphi -\varphi \frac{\partial }{\partial n}\frac{1}{R} \right)$ 
Is going to depend on the boundary conditions.  If you specify the potential on the surface (the $\varphi \frac{\partial }{\partial n}\frac{1}{R}$ term), then this is going to encode the potential from all the charges outside the volume.
If you specify the Electric Field on the surface: (the $\frac{1}{R}\frac{\partial }{\partial n}\varphi $), that also encodes the charges from outside the volume.
This equation doesn't say anything about what the potential is outside the volume.  Only what it is inside.
A: The potential is not zero outside the volume. The source of this confusion is "if the observation point is out side the volume,  $\phi$ is zero" 
This is wrong, really the delta function becomes zero.
