I have a question Inregarding Jackson's Classical Electrodynamics. Consider the jackson electrodynamics , The following equation
$\varphi \left ( x\right )=\tfrac{1}{4\pi\epsilon _{0}} \int \frac{\varrho ( x )}{R}d^{3}x+\tfrac{1}{4\pi} \oint \left(\frac{1}{R}\frac{\partial }{\partial n}\varphi -\varphi \frac{\partial }{\partial n}\frac{1}{R} \right)$$$\varphi \left ( x\right )=\tfrac{1}{4\pi\epsilon _{0}} \int_V \frac{\varrho ( x )}{R}d^{3}x+\tfrac{1}{4\pi} \oint_{\partial V} \left(\frac{1}{R}\frac{\partial }{\partial n}\varphi -\varphi \frac{\partial }{\partial n}\frac{1}{R} \right)dS.$$
Expresses The Electrical PotentialThis expresses the electrostatic potential due to a Charge Distributioncharge distribution $\varrho$ in a Finite Volumefinite volume $V$ with Boundary Conditions Thespecified boundary conditions. The first integral expresses the charge inside a volume in space
. The Secondsecond integral encodes The Boundary Conditionsthe boundary conditions on the surface of this volume ,.
My First Question first question: Doesdoes this determine The Total Potential Completelythe total potential completely inside that Volumevolume regardless Of Any Charge Thatof any charge that you Putput or remove outside The Volumethe volume?Or Or does The Changethe change in The Charge Distribution Outside The Volume Cause Change Of the Boundary Conditionscharge distribution outside the volume cause changes in the boundary conditions (The Potentialthe potential and its normal Derivative Atderivative at the Surfacesurface)?
My Second Question second question:I Can understand Mathematically That The Potential Outside The Volume Should I can understan mathematically that the potential outside the volume should be Zero Butzero but I Can't See Why Thiscan't see why this is So Physically ,Alsoso physically. Also, The Zero potential Outsidethe zero potential outside will Changechange if You Put Someyou put some charge outside the Volume volume, right? Thanks in Advance.
