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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?

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The potential on the outside of the volume can only be zero if the total charge inside the volume is zero. The boundary conditions that determine the second integral are not specified. –  Ron Maimon May 19 '12 at 20:07
    
Ok , But Jackson says in page 37 that the Potential Should be Zero outside the Volume??? –  EGY May 19 '12 at 20:11
3  
I just have to ask, why the randomly capitalized letters? –  Colin K May 20 '12 at 4:41
    
@ColinK Possibly because he's German, and in written German you capitalize all nouns. –  QuantumDot Aug 25 '12 at 23:48
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2 Answers

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.

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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.

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I've edited your answer to reflect what I thought you were saying. Please check to see if it sounds okay now, or you can edit it yourself by clicking "edit" under your answer. –  Kitchi Feb 19 '13 at 13:50
    
Also, you may want to elaborate on what delta function you are talking about. It's quite unclear what exactly you mean. –  Kitchi Feb 19 '13 at 13:51
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